LeetCode-226. 翻转二叉树 + 镜像二叉树 + 递归
题目来源
题目描述
给你一棵二叉树的根节点 root
,翻转这棵二叉树,并返回其根节点。
示例 1:
输入: root = [4,2,7,1,3,6,9]
输出: [4,7,2,9,6,3,1]
示例 2:
输入: root = [2,1,3]
输出: [2,3,1]
示例 3:
输入: root = []
输出: []
提示:
- 树中节点数目范围在
[0, 100]
内 -100 <= Node.val <= 100
题解分析
解法一:后序遍历
- 仔细观察这两颗镜像二叉树,它们的左右子树也分别是一颗镜像二叉树。
- 我们可以利用后序遍历来分别翻转左右子树,最后,我们需要交换root节点的左右子树。
/**
* 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 invertTree(TreeNode root) {
reverse(root);
return root;
}
public void reverse(TreeNode root){
if(root == null)
return;
reverse(root.left);
reverse(root.right);
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
}
}
/**
* 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 invertTree(TreeNode root) {
if(root == null){
return null;
}
TreeNode left = invertTree(root.left);
TreeNode right = invertTree(root.right);
root.left = right;
root.right = left;
return root;
}
}
Either Excellent or Rusty