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374 lines
11 KiB
Markdown
374 lines
11 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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> 利用二叉搜索树的特性搞起！
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# 530.二叉搜索树的最小绝对差
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[力扣题目链接](https://leetcode-cn.com/problems/minimum-absolute-difference-in-bst/)
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给你一棵所有节点为非负值的二叉搜索树，请你计算树中任意两节点的差的绝对值的最小值。
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示例：
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提示：树中至少有 2 个节点。
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# 思路
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题目中要求在二叉搜索树上任意两节点的差的绝对值的最小值。
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**注意是二叉搜索树**，二叉搜索树可是有序的。
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遇到在二叉搜索树上求什么最值啊，差值之类的，就把它想成在一个有序数组上求最值，求差值，这样就简单多了。
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## 递归
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那么二叉搜索树采用中序遍历，其实就是一个有序数组。
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**在一个有序数组上求两个数最小差值，这是不是就是一道送分题了。**
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最直观的想法，就是把二叉搜索树转换成有序数组，然后遍历一遍数组，就统计出来最小差值了。
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代码如下：
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```CPP
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class Solution {
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private:
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vector<int> vec;
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void traversal(TreeNode* root) {
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if (root == NULL) return;
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traversal(root->left);
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vec.push_back(root->val); // 将二叉搜索树转换为有序数组
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traversal(root->right);
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}
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public:
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int getMinimumDifference(TreeNode* root) {
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vec.clear();
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traversal(root);
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if (vec.size() < 2) return 0;
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int result = INT_MAX;
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for (int i = 1; i < vec.size(); i++) { // 统计有序数组的最小差值
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result = min(result, vec[i] - vec[i-1]);
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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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以上代码是把二叉搜索树转化为有序数组了，其实在二叉搜素树中序遍历的过程中，我们就可以直接计算了。
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需要用一个pre节点记录一下cur节点的前一个节点。
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如图：
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一些同学不知道在递归中如何记录前一个节点的指针，其实实现起来是很简单的，大家只要看过一次，写过一次，就掌握了。
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代码如下：
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```CPP
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class Solution {
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private:
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int result = INT_MAX;
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TreeNode* pre;
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void traversal(TreeNode* cur) {
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if (cur == NULL) return;
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traversal(cur->left); // 左
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if (pre != NULL){ // 中
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result = min(result, cur->val - pre->val);
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}
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pre = cur; // 记录前一个
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traversal(cur->right); // 右
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}
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public:
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int getMinimumDifference(TreeNode* root) {
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traversal(root);
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return result;
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}
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};
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```
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是不是看上去也并不复杂！
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## 迭代
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看过这两篇[二叉树：听说递归能做的，栈也能做！](https://programmercarl.com/二叉树的迭代遍历.html)，[二叉树：前中后序迭代方式的写法就不能统一一下么？](https://programmercarl.com/二叉树的统一迭代法.html)文章之后，不难写出两种中序遍历的迭代法。
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下面我给出其中的一种中序遍历的迭代法，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 getMinimumDifference(TreeNode* root) {
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stack<TreeNode*> st;
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TreeNode* cur = root;
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TreeNode* pre = NULL;
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int result = INT_MAX;
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while (cur != NULL || !st.empty()) {
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if (cur != NULL) { // 指针来访问节点，访问到最底层
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st.push(cur); // 将访问的节点放进栈
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cur = cur->left; // 左
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} else {
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cur = st.top();
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st.pop();
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if (pre != NULL) { // 中
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result = min(result, cur->val - pre->val);
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}
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pre = cur;
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cur = cur->right; // 右
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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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# 总结
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**遇到在二叉搜索树上求什么最值，求差值之类的，都要思考一下二叉搜索树可是有序的，要利用好这一特点。**
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同时要学会在递归遍历的过程中如何记录前后两个指针，这也是一个小技巧，学会了还是很受用的。
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后面我将继续介绍一系列利用二叉搜索树特性的题目。
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# 其他语言版本
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## Java
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递归
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```java
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class Solution {
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TreeNode pre;// 记录上一个遍历的结点
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int result = Integer.MAX_VALUE;
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public int getMinimumDifference(TreeNode root) {
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if(root==null)return 0;
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traversal(root);
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return result;
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}
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public void traversal(TreeNode root){
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if(root==null)return;
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//左
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traversal(root.left);
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//中
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if(pre!=null){
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result = Math.min(result,root.val-pre.val);
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}
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pre = root;
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//右
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traversal(root.right);
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}
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}
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```
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迭代法-中序遍历
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```java
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class Solution {
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TreeNode pre;
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Stack<TreeNode> stack;
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public int getMinimumDifference(TreeNode root) {
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if (root == null) return 0;
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stack = new Stack<>();
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TreeNode cur = root;
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int result = Integer.MAX_VALUE;
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while (cur != null || !stack.isEmpty()) {
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if (cur != null) {
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stack.push(cur); // 将访问的节点放进栈
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cur = cur.left; // 左
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}else {
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cur = stack.pop();
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if (pre != null) { // 中
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result = Math.min(result, cur.val - pre.val);
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}
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pre = cur;
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cur = cur.right; // 右
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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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## Python
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递归
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```python
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class Solution:
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def getMinimumDifference(self, root: TreeNode) -> int:
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res = []
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r = float("inf")
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def buildaList(root): //把二叉搜索树转换成有序数组
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if not root: return None
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if root.left: buildaList(root.left) //左
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res.append(root.val) //中
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if root.right: buildaList(root.right) //右
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return res
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buildaList(root)
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for i in range(len(res)-1): // 统计有序数组的最小差值
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r = min(abs(res[i]-res[i+1]),r)
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return r
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```
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迭代法-中序遍历
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```python
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class Solution:
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def getMinimumDifference(self, root: TreeNode) -> int:
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stack = []
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cur = root
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pre = None
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result = float('inf')
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while cur or stack:
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if cur: # 指针来访问节点，访问到最底层
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stack.append(cur)
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cur = cur.left
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else: # 逐一处理节点
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cur = stack.pop()
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if pre: # 当前节点和前节点的值的差值
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result = min(result, cur.val - pre.val)
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pre = cur
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cur = cur.right
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return result
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```
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## Go：
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中序遍历，然后计算最小差值
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```go
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func getMinimumDifference(root *TreeNode) int {
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var res []int
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findMIn(root,&res)
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min:=1000000//一个比较大的值
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for i:=1;i<len(res);i++{
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tempValue:=res[i]-res[i-1]
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if tempValue<min{
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min=tempValue
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}
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}
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return min
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}
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//中序遍历
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func findMIn(root *TreeNode,res *[]int){
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if root==nil{return}
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findMIn(root.Left,res)
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*res=append(*res,root.Val)
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findMIn(root.Right,res)
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}
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```
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```go
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// 中序遍历的同时计算最小值
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func getMinimumDifference(root *TreeNode) int {
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// 保留前一个节点的指针
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var prev *TreeNode
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// 定义一个比较大的值
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min := math.MaxInt64
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var travel func(node *TreeNode)
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travel = func(node *TreeNode) {
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if node == nil {
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return
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}
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travel(node.Left)
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if prev != nil && node.Val - prev.Val < min {
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min = node.Val - prev.Val
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}
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prev = node
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travel(node.Right)
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}
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travel(root)
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return min
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}
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```
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## JavaScript
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递归 先转换为有序数组
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```javascript
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/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {number}
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*/
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var getMinimumDifference = function (root) {
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let arr = [];
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const buildArr = (root) => {
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if (root) {
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buildArr(root.left);
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arr.push(root.val);
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buildArr(root.right);
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}
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}
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buildArr(root);
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let diff = arr[arr.length - 1];
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for (let i = 1; i < arr.length; ++i) {
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if (diff > arr[i] - arr[i - 1])
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diff = arr[i] - arr[i - 1];
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}
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return diff;
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};
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```
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递归 在递归的过程中更新最小值
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```js
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var getMinimumDifference = function(root) {
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let res = Infinity
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let preNode = null
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// 中序遍历
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const inorder = (node) => {
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if(!node) return
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inorder(node.left)
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// 更新res
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if(preNode) res = Math.min(res, node.val - preNode.val)
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// 记录前一个节点
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preNode = node
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inorder(node.right)
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}
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inorder(root)
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return res
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}
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```
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迭代 中序遍历
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```js
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var getMinimumDifference = function(root) {
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let stack = []
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let cur = root
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let res = Infinity
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let pre = null
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while(cur || stack.length) {
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if(cur) {
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stack.push(cur)
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cur = cur.left
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} else {
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cur = stack.pop()
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if(pre) res = Math.min(res, cur.val - pre.val)
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pre = cur
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cur = cur.right
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}
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}
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return res
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}
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```
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-----------------------
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* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频：[代码随想录](https://space.bilibili.com/525438321)
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* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>
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