【LeetCode笔记】226. 翻转二叉树（Java、递归）

题目描述

• 感觉和合并二叉树类似，都是很好进行递归的问题
  

代码 & 思路

• 翻转当前结点的左、右结点
• 对当前结点的左、右结点进行翻转函数
• 【自底向上】

/**
 * 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 {
    TreeNode left;
    public TreeNode invertTree(TreeNode root) {
        if(root == null){
            return root;
        }
        invertTree(root.left);
        invertTree(root.right);
        left = root.left;
        root.left = root.right;
        root.right = left;
        return root;
    }
}

更新版

• 自顶向下，感觉这个比较好理解

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root == null) {
            return null;
        }
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}

• 面试前再刷一次，属实是过于清晰了

class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root == null) return null;
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        mirrorTree(root.left);
        mirrorTree(root.right);
        return root;
    }
}