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645 lines
17 KiB
Markdown
645 lines
17 KiB
Markdown
<p align="center">
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<a href="https://mp.weixin.qq.com/s/RsdcQ9umo09R6cfnwXZlrQ"><img src="https://img.shields.io/badge/PDF下载-代码随想录-blueviolet" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw"><img src="https://img.shields.io/badge/刷题-微信群-green" alt=""></a>
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<a href="https://space.bilibili.com/525438321"><img src="https://img.shields.io/badge/B站-代码随想录-orange" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
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</p>
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<p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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看完本篇可以一起做了如下两道题目：
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* 104.二叉树的最大深度
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* 559.n叉树的最大深度
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# 104.二叉树的最大深度
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[力扣题目链接](https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/)
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给定一个二叉树，找出其最大深度。
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二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
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说明: 叶子节点是指没有子节点的节点。
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示例：
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给定二叉树 [3,9,20,null,null,15,7]，
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返回它的最大深度 3 。
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## 递归法
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本题可以使用前序（中左右），也可以使用后序遍历（左右中），使用前序求的就是深度，使用后序求的是高度。
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**而根节点的高度就是二叉树的最大深度**，所以本题中我们通过后序求的根节点高度来求的二叉树最大深度。
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这一点其实是很多同学没有想清楚的，很多题解同样没有讲清楚。
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我先用后序遍历（左右中）来计算树的高度。
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1. 确定递归函数的参数和返回值：参数就是传入树的根节点，返回就返回这棵树的深度，所以返回值为int类型。
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代码如下：
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```CPP
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int getdepth(treenode* node)
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```
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2. 确定终止条件：如果为空节点的话，就返回0，表示高度为0。
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代码如下：
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```CPP
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if (node == null) return 0;
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```
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3. 确定单层递归的逻辑：先求它的左子树的深度，再求的右子树的深度，最后取左右深度最大的数值 再+1 （加1是因为算上当前中间节点）就是目前节点为根节点的树的深度。
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代码如下：
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```CPP
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int leftdepth = getdepth(node->left); // 左
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int rightdepth = getdepth(node->right); // 右
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int depth = 1 + max(leftdepth, rightdepth); // 中
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return depth;
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```
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所以整体c++代码如下：
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```CPP
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class solution {
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public:
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int getdepth(treenode* node) {
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if (node == null) return 0;
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int leftdepth = getdepth(node->left); // 左
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int rightdepth = getdepth(node->right); // 右
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int depth = 1 + max(leftdepth, rightdepth); // 中
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return depth;
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}
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int maxdepth(treenode* root) {
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return getdepth(root);
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}
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};
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```
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代码精简之后c++代码如下：
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```CPP
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class solution {
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public:
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int maxdepth(treenode* root) {
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if (root == null) return 0;
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return 1 + max(maxdepth(root->left), maxdepth(root->right));
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}
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};
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```
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**精简之后的代码根本看不出是哪种遍历方式，也看不出递归三部曲的步骤，所以如果对二叉树的操作还不熟练，尽量不要直接照着精简代码来学。**
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本题当然也可以使用前序，代码如下：(**充分表现出求深度回溯的过程**)
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```CPP
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class solution {
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public:
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int result;
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void getdepth(treenode* node, int depth) {
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result = depth > result ? depth : result; // 中
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if (node->left == null && node->right == null) return ;
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if (node->left) { // 左
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depth++; // 深度+1
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getdepth(node->left, depth);
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depth--; // 回溯，深度-1
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}
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if (node->right) { // 右
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depth++; // 深度+1
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getdepth(node->right, depth);
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depth--; // 回溯，深度-1
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}
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return ;
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}
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int maxdepth(treenode* root) {
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result = 0;
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if (root == NULL) return result;
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getdepth(root, 1);
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return result;
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}
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};
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```
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**可以看出使用了前序（中左右）的遍历顺序，这才是真正求深度的逻辑！**
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注意以上代码是为了把细节体现出来，简化一下代码如下：
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```CPP
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class solution {
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public:
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int result;
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void getdepth(treenode* node, int depth) {
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result = depth > result ? depth : result; // 中
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if (node->left == null && node->right == null) return ;
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if (node->left) { // 左
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getdepth(node->left, depth + 1);
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}
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if (node->right) { // 右
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getdepth(node->right, depth + 1);
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}
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return ;
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}
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int maxdepth(treenode* root) {
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result = 0;
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if (root == 0) return result;
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getdepth(root, 1);
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return result;
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}
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};
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```
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## 迭代法
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使用迭代法的话，使用层序遍历是最为合适的，因为最大的深度就是二叉树的层数，和层序遍历的方式极其吻合。
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在二叉树中，一层一层的来遍历二叉树，记录一下遍历的层数就是二叉树的深度，如图所示：
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所以这道题的迭代法就是一道模板题，可以使用二叉树层序遍历的模板来解决的。
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如果对层序遍历还不清楚的话，可以看这篇：[二叉树：层序遍历登场！](https://programmercarl.com/0102.二叉树的层序遍历.html)
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c++代码如下：
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```CPP
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class solution {
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public:
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int maxdepth(treenode* root) {
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if (root == null) return 0;
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int depth = 0;
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queue<treenode*> que;
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que.push(root);
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while(!que.empty()) {
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int size = que.size();
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depth++; // 记录深度
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for (int i = 0; i < size; i++) {
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treenode* node = que.front();
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que.pop();
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if (node->left) que.push(node->left);
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if (node->right) que.push(node->right);
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}
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}
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return depth;
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}
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};
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```
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那么我们可以顺便解决一下n叉树的最大深度问题
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# 559.n叉树的最大深度
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[力扣题目链接](https://leetcode-cn.com/problems/maximum-depth-of-n-ary-tree/)
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给定一个 n 叉树，找到其最大深度。
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最大深度是指从根节点到最远叶子节点的最长路径上的节点总数。
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例如，给定一个 3叉树 :
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我们应返回其最大深度，3。
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思路：
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依然可以提供递归法和迭代法，来解决这个问题，思路是和二叉树思路一样的，直接给出代码如下：
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## 递归法
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c++代码：
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```CPP
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class solution {
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public:
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int maxdepth(node* root) {
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if (root == 0) return 0;
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int depth = 0;
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for (int i = 0; i < root->children.size(); i++) {
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depth = max (depth, maxdepth(root->children[i]));
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}
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return depth + 1;
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}
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};
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```
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## 迭代法
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依然是层序遍历，代码如下：
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```CPP
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class solution {
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public:
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int maxdepth(node* root) {
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queue<node*> que;
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if (root != null) que.push(root);
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int depth = 0;
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while (!que.empty()) {
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int size = que.size();
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depth++; // 记录深度
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for (int i = 0; i < size; i++) {
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node* node = que.front();
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que.pop();
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for (int j = 0; j < node->children.size(); j++) {
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if (node->children[j]) que.push(node->children[j]);
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}
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}
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}
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return depth;
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}
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};
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```
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# 其他语言版本
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## java
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### 104.二叉树的最大深度
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```java
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class solution {
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/**
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* 递归法
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*/
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public int maxdepth(treenode root) {
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if (root == null) {
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return 0;
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}
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int leftdepth = maxdepth(root.left);
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int rightdepth = maxdepth(root.right);
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return math.max(leftdepth, rightdepth) + 1;
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}
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}
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```
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```java
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class solution {
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/**
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* 迭代法，使用层序遍历
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*/
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public int maxdepth(treenode root) {
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if(root == null) {
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return 0;
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}
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deque<treenode> deque = new linkedlist<>();
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deque.offer(root);
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int depth = 0;
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while (!deque.isempty()) {
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int size = deque.size();
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depth++;
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for (int i = 0; i < size; i++) {
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treenode poll = deque.poll();
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if (poll.left != null) {
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deque.offer(poll.left);
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}
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if (poll.right != null) {
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deque.offer(poll.right);
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}
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}
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}
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return depth;
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}
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}
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```
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### 559.n叉树的最大深度
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```java
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class solution {
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/**
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* 迭代法，使用层序遍历
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*/
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public int maxDepth(Node root) {
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if (root == null) return 0;
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int depth = 0;
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Queue<Node> que = new LinkedList<>();
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que.offer(root);
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while (!que.isEmpty())
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{
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depth ++;
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int len = que.size();
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while (len > 0)
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{
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Node node = que.poll();
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for (int i = 0; i < node.children.size(); i++)
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if (node.children.get(i) != null)
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que.offer(node.children.get(i));
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len--;
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}
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}
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return depth;
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}
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}
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```
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## python
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### 104.二叉树的最大深度
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递归法：
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```python
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class solution:
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def maxdepth(self, root: treenode) -> int:
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return self.getdepth(root)
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def getdepth(self, node):
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if not node:
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return 0
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leftdepth = self.getdepth(node.left) #左
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rightdepth = self.getdepth(node.right) #右
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depth = 1 + max(leftdepth, rightdepth) #中
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return depth
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```
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递归法：精简代码
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```python
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class solution:
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def maxdepth(self, root: treenode) -> int:
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if not root:
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return 0
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return 1 + max(self.maxdepth(root.left), self.maxdepth(root.right))
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```
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迭代法：
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```python
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import collections
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class solution:
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def maxdepth(self, root: treenode) -> int:
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if not root:
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return 0
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depth = 0 #记录深度
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queue = collections.deque()
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queue.append(root)
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while queue:
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size = len(queue)
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depth += 1
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for i in range(size):
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node = queue.popleft()
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if node.left:
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queue.append(node.left)
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if node.right:
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queue.append(node.right)
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return depth
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```
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### 559.n叉树的最大深度
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递归法：
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```python
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class solution:
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def maxdepth(self, root: 'node') -> int:
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if not root:
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return 0
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depth = 0
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for i in range(len(root.children)):
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depth = max(depth, self.maxdepth(root.children[i]))
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return depth + 1
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```
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迭代法：
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```python
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import collections
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class solution:
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def maxdepth(self, root: 'node') -> int:
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queue = collections.deque()
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if root:
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queue.append(root)
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depth = 0 #记录深度
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while queue:
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size = len(queue)
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depth += 1
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for i in range(size):
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node = queue.popleft()
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for j in range(len(node.children)):
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if node.children[j]:
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queue.append(node.children[j])
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return depth
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```
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使用栈来模拟后序遍历依然可以
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```python
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class solution:
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def maxdepth(self, root: 'node') -> int:
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st = []
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if root:
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st.append(root)
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depth = 0
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result = 0
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while st:
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node = st.pop()
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if node != none:
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st.append(node) #中
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st.append(none)
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depth += 1
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for i in range(len(node.children)): #处理孩子
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if node.children[i]:
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st.append(node.children[i])
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else:
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node = st.pop()
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depth -= 1
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result = max(result, depth)
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return result
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```
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## go
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```go
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/**
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* definition for a binary tree node.
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* type treenode struct {
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* val int
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* left *treenode
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* right *treenode
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* }
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*/
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func max (a, b int) int {
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if a > b {
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return a;
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}
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return b;
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}
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// 递归
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func maxdepth(root *treenode) int {
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if root == nil {
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return 0;
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}
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return max(maxdepth(root.left), maxdepth(root.right)) + 1;
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}
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// 遍历
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func maxdepth(root *treenode) int {
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levl := 0;
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queue := make([]*treenode, 0);
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if root != nil {
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queue = append(queue, root);
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}
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for l := len(queue); l > 0; {
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for ;l > 0;l-- {
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node := queue[0];
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if node.left != nil {
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queue = append(queue, node.left);
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}
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if node.right != nil {
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queue = append(queue, node.right);
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}
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queue = queue[1:];
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}
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levl++;
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l = len(queue);
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}
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return levl;
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}
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```
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## javascript
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```javascript
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var maxdepth = function(root) {
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if (!root) return root
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return 1 + math.max(maxdepth(root.left), maxdepth(root.right))
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};
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```
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二叉树最大深度递归遍历
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```javascript
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var maxdepth = function(root) {
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//使用递归的方法 递归三部曲
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//1. 确定递归函数的参数和返回值
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const getdepth=function(node){
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//2. 确定终止条件
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if(node===null){
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return 0;
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}
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//3. 确定单层逻辑
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let leftdepth=getdepth(node.left);
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let rightdepth=getdepth(node.right);
|
||
let depth=1+math.max(leftdepth,rightdepth);
|
||
return depth;
|
||
}
|
||
return getdepth(root);
|
||
};
|
||
```
|
||
|
||
二叉树最大深度层级遍历
|
||
```javascript
|
||
var maxDepth = function(root) {
|
||
if(!root) return 0
|
||
let count = 0
|
||
const queue = [root]
|
||
while(queue.length) {
|
||
let size = queue.length
|
||
/* 层数+1 */
|
||
count++
|
||
while(size--) {
|
||
let node = queue.shift();
|
||
node.left && queue.push(node.left);
|
||
node.right && queue.push(node.right);
|
||
}
|
||
}
|
||
return count
|
||
};
|
||
```
|
||
|
||
N叉树的最大深度 递归写法
|
||
```js
|
||
var maxDepth = function(root) {
|
||
if(!root) return 0
|
||
let depth = 0
|
||
for(let node of root.children) {
|
||
depth = Math.max(depth, maxDepth(node))
|
||
}
|
||
return depth + 1
|
||
}
|
||
```
|
||
|
||
N叉树的最大深度 层序遍历
|
||
```js
|
||
var maxDepth = function(root) {
|
||
if(!root) return 0
|
||
let count = 0
|
||
let queue = [root]
|
||
while(queue.length) {
|
||
let size = queue.length
|
||
count++
|
||
while(size--) {
|
||
let node = queue.shift()
|
||
node && (queue = [...queue, ...node.children])
|
||
}
|
||
}
|
||
return count
|
||
};
|
||
```
|
||
|
||
## C
|
||
二叉树最大深度递归
|
||
```c
|
||
int maxDepth(struct TreeNode* root){
|
||
//若传入结点为NULL，返回0
|
||
if(!root)
|
||
return 0;
|
||
|
||
//求出左子树深度
|
||
int left = maxDepth(root->left);
|
||
//求出右子树深度
|
||
int right = maxDepth(root->right);
|
||
//求出左子树深度和右子树深度的较大值
|
||
int max = left > right ? left : right;
|
||
//返回较大值+1（1为当前层数）
|
||
return max + 1;
|
||
}
|
||
```
|
||
二叉树最大深度迭代
|
||
```c
|
||
int maxDepth(struct TreeNode* root){
|
||
//若传入根节点为NULL，返回0
|
||
if(!root)
|
||
return 0;
|
||
|
||
int depth = 0;
|
||
//开辟队列空间
|
||
struct TreeNode** queue = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 6000);
|
||
int queueFront = 0;
|
||
int queueEnd = 0;
|
||
|
||
//将根结点入队
|
||
queue[queueEnd++] = root;
|
||
|
||
int queueSize;
|
||
//求出当前队列中元素个数
|
||
while(queueSize = queueEnd - queueFront) {
|
||
int i;
|
||
//若当前队列中结点有左右子树，则将它们的左右子树入队
|
||
for(i = 0; i < queueSize; i++) {
|
||
struct TreeNode* tempNode = queue[queueFront + i];
|
||
if(tempNode->left)
|
||
queue[queueEnd++] = tempNode->left;
|
||
if(tempNode->right)
|
||
queue[queueEnd++] = tempNode->right;
|
||
}
|
||
//更新队头下标
|
||
queueFront += queueSize;
|
||
//深度+1
|
||
depth++;
|
||
}
|
||
return depth;
|
||
}
|
||
```
|
||
|
||
-----------------------
|
||
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
|
||
* B站视频：[代码随想录](https://space.bilibili.com/525438321)
|
||
* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
|
||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|