LeetCode450题—— 删除二叉搜索树中的节点

在这里插入图片描述

首先需要认识什么是二叉搜索树，可以进入百度词条https://baike.baidu.com/item/%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91/7077855?fr=aladdin

一定要注意题目要求算法时间复杂度是O(h)。

递归方法：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public TreeNode deleteNode(TreeNode root, int key) {if (root == null)return null;if (key == root.val) {// 1.leafif (root.left == null && root.right == null) {return null;} else if (root.left == null) { // 2.left is nullreturn root.right;  } else if (root.right == null) { // 3.right is nullreturn root.left;} else { // 4.left and right are not nullTreeNode rightTreeMin = root.right;while (rightTreeMin.left != null) {rightTreeMin = rightTreeMin.left; //找到右子树中最小的叶子节点}root.val = rightTreeMin.val;root.right = deleteNode(root.right, rightTreeMin.val);//继续向当前root节点的右子树递归，此时的key变成了rightTreeMin.val，以达到去删除这个最小叶子节点的目的}} else if (key > root.val) {root.right = deleteNode(root.right, key);} else {   //key<root.valroot.left = deleteNode(root.left, key);}return root;}
}