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532 lines
15 KiB
Markdown
532 lines
15 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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# 222.完全二叉树的节点个数
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[力扣题目链接](https://leetcode-cn.com/problems/count-complete-tree-nodes/)
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给出一个完全二叉树，求出该树的节点个数。
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示例 1：
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* 输入：root = [1,2,3,4,5,6]
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* 输出：6
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示例 2：
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* 输入：root = []
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* 输出：0
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示例 3：
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* 输入：root = [1]
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* 输出：1
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提示：
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* 树中节点的数目范围是[0, 5 * 10^4]
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* 0 <= Node.val <= 5 * 10^4
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* 题目数据保证输入的树是 完全二叉树
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# 思路
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本篇给出按照普通二叉树的求法以及利用完全二叉树性质的求法。
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## 普通二叉树
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首先按照普通二叉树的逻辑来求。
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这道题目的递归法和求二叉树的深度写法类似， 而迭代法，[二叉树：层序遍历登场！](https://programmercarl.com/0102.二叉树的层序遍历.html)遍历模板稍稍修改一下，记录遍历的节点数量就可以了。
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递归遍历的顺序依然是后序（左右中）。
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### 递归
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如果对求二叉树深度还不熟悉的话，看这篇：[二叉树：看看这些树的最大深度](https://programmercarl.com/0104.二叉树的最大深度.html)。
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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，返回就返回以该节点为根节点二叉树的节点数量，所以返回值为int类型。
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代码如下：
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```
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int getNodesNum(TreeNode* cur) {
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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，表示节点数为0。
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代码如下：
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```
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if (cur == NULL) return 0;
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```
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3. 确定单层递归的逻辑：先求它的左子树的节点数量，再求的右子树的节点数量，最后取总和再加一 （加1是因为算上当前中间节点）就是目前节点为根节点的节点数量。
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代码如下：
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```
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int leftNum = getNodesNum(cur->left); // 左
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int rightNum = getNodesNum(cur->right); // 右
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int treeNum = leftNum + rightNum + 1; // 中
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return treeNum;
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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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// 版本一
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class Solution {
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private:
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int getNodesNum(TreeNode* cur) {
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if (cur == 0) return 0;
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int leftNum = getNodesNum(cur->left); // 左
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int rightNum = getNodesNum(cur->right); // 右
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int treeNum = leftNum + rightNum + 1; // 中
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return treeNum;
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}
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public:
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int countNodes(TreeNode* root) {
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return getNodesNum(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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// 版本二
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class Solution {
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public:
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int countNodes(TreeNode* root) {
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if (root == NULL) return 0;
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return 1 + countNodes(root->left) + countNodes(root->right);
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}
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};
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```
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* 时间复杂度：O(n)
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* 空间复杂度：O(logn)，算上了递归系统栈占用的空间
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**网上基本都是这个精简的代码版本，其实不建议大家照着这个来写，代码确实精简，但隐藏了一些内容，连遍历的顺序都看不出来，所以初学者建议学习版本一的代码，稳稳的打基础**。
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### 迭代法
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如果对求二叉树层序遍历还不熟悉的话，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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那么只要模板少做改动，加一个变量result，统计节点数量就可以了
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```CPP
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class Solution {
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public:
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int countNodes(TreeNode* root) {
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queue<TreeNode*> que;
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if (root != NULL) que.push(root);
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int result = 0;
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while (!que.empty()) {
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int size = que.size();
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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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result++; // 记录节点数量
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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 result;
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}
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};
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```
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* 时间复杂度：O(n)
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* 空间复杂度：O(n)
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## 完全二叉树
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以上方法都是按照普通二叉树来做的，对于完全二叉树特性不了解的同学可以看这篇 [关于二叉树，你该了解这些！](https://programmercarl.com/二叉树理论基础.html)，这篇详细介绍了各种二叉树的特性。
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完全二叉树只有两种情况，情况一：就是满二叉树，情况二：最后一层叶子节点没有满。
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对于情况一，可以直接用 2^树深度 - 1 来计算，注意这里根节点深度为1。
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对于情况二，分别递归左孩子，和右孩子，递归到某一深度一定会有左孩子或者右孩子为满二叉树，然后依然可以按照情况1来计算。
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完全二叉树（一）如图：
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完全二叉树（二）如图：
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可以看出如果整个树不是满二叉树，就递归其左右孩子，直到遇到满二叉树为止，用公式计算这个子树（满二叉树）的节点数量。
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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 countNodes(TreeNode* root) {
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if (root == nullptr) return 0;
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TreeNode* left = root->left;
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TreeNode* right = root->right;
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int leftHeight = 0, rightHeight = 0; // 这里初始为0是有目的的，为了下面求指数方便
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while (left) { // 求左子树深度
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left = left->left;
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leftHeight++;
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}
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while (right) { // 求右子树深度
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right = right->right;
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rightHeight++;
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}
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if (leftHeight == rightHeight) {
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return (2 << leftHeight) - 1; // 注意(2<<1) 相当于2^2，所以leftHeight初始为0
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}
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return countNodes(root->left) + countNodes(root->right) + 1;
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}
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};
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```
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* 时间复杂度：O(logn * logn)
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* 空间复杂度：O(logn)
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# 其他语言版本
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## Java
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```java
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class Solution {
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// 通用递归解法
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public int countNodes(TreeNode root) {
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if(root == null) {
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return 0;
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}
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return countNodes(root.left) + countNodes(root.right) + 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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// 迭代法
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public int countNodes(TreeNode root) {
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if (root == null) return 0;
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Queue<TreeNode> queue = new LinkedList<>();
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queue.offer(root);
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int result = 0;
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while (!queue.isEmpty()) {
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int size = queue.size();
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while (size -- > 0) {
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TreeNode cur = queue.poll();
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result++;
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if (cur.left != null) queue.offer(cur.left);
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if (cur.right != null) queue.offer(cur.right);
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}
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}
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return result;
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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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* 针对完全二叉树的解法
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*
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* 满二叉树的结点数为：2^depth - 1
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*/
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public int countNodes(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 = getDepth(root.left);
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int rightDepth = getDepth(root.right);
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if (leftDepth == rightDepth) {// 左子树是满二叉树
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// 2^leftDepth其实是 （2^leftDepth - 1） + 1 ，左子树 + 根结点
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return (1 << leftDepth) + countNodes(root.right);
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} else {// 右子树是满二叉树
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return (1 << rightDepth) + countNodes(root.left);
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}
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}
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private int getDepth(TreeNode root) {
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int depth = 0;
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while (root != null) {
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root = root.left;
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depth++;
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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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递归法：
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```python
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class Solution:
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def countNodes(self, root: TreeNode) -> int:
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return self.getNodesNum(root)
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def getNodesNum(self, cur):
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if not cur:
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return 0
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leftNum = self.getNodesNum(cur.left) #左
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rightNum = self.getNodesNum(cur.right) #右
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treeNum = leftNum + rightNum + 1 #中
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return treeNum
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```
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递归法：精简版
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```python
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class Solution:
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def countNodes(self, root: TreeNode) -> int:
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if not root:
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return 0
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return 1 + self.countNodes(root.left) + self.countNodes(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 countNodes(self, root: TreeNode) -> 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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result = 0
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while queue:
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size = len(queue)
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for i in range(size):
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node = queue.popleft()
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result += 1 #记录节点数量
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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 result
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```
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完全二叉树
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```python
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class Solution:
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def countNodes(self, root: TreeNode) -> int:
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if not root:
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return 0
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left = root.left
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right = root.right
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leftHeight = 0 #这里初始为0是有目的的，为了下面求指数方便
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rightHeight = 0
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while left: #求左子树深度
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left = left.left
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leftHeight += 1
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while right: #求右子树深度
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right = right.right
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rightHeight += 1
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if leftHeight == rightHeight:
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return (2 << leftHeight) - 1 #注意(2<<1) 相当于2^2，所以leftHeight初始为0
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return self.countNodes(root.left) + self.countNodes(root.right) + 1
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```
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## Go
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递归版本
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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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//本题直接就是求有多少个节点，无脑存进数组算长度就行了。
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func countNodes(root *TreeNode) int {
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if root == nil {
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return 0
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}
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res := 1
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if root.Right != nil {
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res += countNodes(root.Right)
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}
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if root.Left != nil {
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res += countNodes(root.Left)
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}
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return res
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}
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```
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利用完全二叉树特性的递归解法
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```go
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func countNodes(root *TreeNode) int {
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if root == nil {
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return 0
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}
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leftH, rightH := 0, 0
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leftNode := root.Left
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rightNode := root.Right
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for leftNode != nil {
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leftNode = leftNode.Left
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leftH++
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}
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for rightNode != nil {
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rightNode = rightNode.Right
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rightH++
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}
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if leftH == rightH {
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return (2 << leftH) - 1
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}
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return countNodes(root.Left) + countNodes(root.Right) + 1
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}
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```
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## JavaScript:
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递归版本
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```javascript
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var countNodes = function(root) {
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//递归法计算二叉树节点数
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// 1. 确定递归函数参数
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const getNodeSum=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. 确定单层递归逻辑
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let leftNum=getNodeSum(node.left);
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let rightNum=getNodeSum(node.right);
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return leftNum+rightNum+1;
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}
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return getNodeSum(root);
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};
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```
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迭代(层序遍历)版本
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```javascript
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var countNodes = function(root) {
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//层序遍历
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let queue=[];
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if(root===null){
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return 0;
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}
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queue.push(root);
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let nodeNums=0;
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while(queue.length){
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let length=queue.length;
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while(length--){
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let node=queue.shift();
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nodeNums++;
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node.left&&queue.push(node.left);
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node.right&&queue.push(node.right);
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}
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}
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return nodeNums;
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};
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```
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利用完全二叉树性质
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```javascript
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var countNodes = function(root) {
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//利用完全二叉树的特点
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if(root===null){
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return 0;
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}
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let left=root.left;
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let right=root.right;
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let leftHeight=0,rightHeight=0;
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while(left){
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left=left.left;
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leftHeight++;
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}
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while(right){
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right=right.right;
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rightHeight++;
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}
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if(leftHeight==rightHeight){
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return Math.pow(2,leftHeight+1)-1;
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}
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return countNodes(root.left)+countNodes(root.right)+1;
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};
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```
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## C:
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递归法
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```c
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int countNodes(struct TreeNode* root) {
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//若传入结点不存在，返回0
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if(!root)
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return 0;
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//算出左右子树的结点总数
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int leftCount = countNodes(root->left);
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int rightCount = countNodes(root->right);
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//返回左右子树结点总数+1
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return leftCount + rightCount + 1;
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}
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int countNodes(struct TreeNode* root){
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return getNodes(root);
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}
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```
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迭代法
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```c
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int countNodes(struct TreeNode* root){
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//记录结点总数
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int totalNum = 0;
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//开辟栈空间
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struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 100);
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int stackTop = 0;
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//如果root结点不为NULL，则将其入栈。若为NULL，则不会进入遍历，返回0
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if(root)
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stack[stackTop++] = root;
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//若栈中有结点存在，则进行遍历
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while(stackTop) {
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//取出栈顶元素
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struct TreeNode* tempNode = stack[--stackTop];
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//结点总数+1
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totalNum++;
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//若栈顶结点有左右孩子，将它们入栈
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||
if(tempNode->left)
|
||
stack[stackTop++] = tempNode->left;
|
||
if(tempNode->right)
|
||
stack[stackTop++] = tempNode->right;
|
||
}
|
||
return totalNum;
|
||
}
|
||
```
|
||
|
||
满二叉树
|
||
```c
|
||
int countNodes(struct TreeNode* root){
|
||
if(!root)
|
||
return 0;
|
||
int leftHeight = 0;
|
||
int rightHeight = 0;
|
||
struct TreeNode* rightNode = root->right;
|
||
struct TreeNode* leftNode = root->left;
|
||
//求出左子树深度
|
||
while(leftNode) {
|
||
leftNode = leftNode->left;
|
||
leftHeight++;
|
||
}
|
||
|
||
//求出右子树深度
|
||
while(rightNode) {
|
||
rightNode = rightNode->right;
|
||
rightHeight++;
|
||
}
|
||
//若左右子树深度相同，为满二叉树。结点个数为2^height-1
|
||
if(rightHeight == leftHeight) {
|
||
return (2 << leftHeight) - 1;
|
||
}
|
||
//否则返回左右子树的结点个数+1
|
||
return countNodes(root->right) + countNodes(root->left) + 1;
|
||
}
|
||
```
|
||
|
||
-----------------------
|
||
* 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>
|