力扣热题100题-翻转二叉树


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:

img

输入:root = [4,2,7,1,3,6,9]
输出:[4,7,2,9,6,3,1]
示例 2:

img

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

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/invert-binary-tree

思路:

/**
 * 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) {
        //dayinroot
        if(root == null){
            return null ;
        }
        TreeNode result = new TreeNode(root.val);
        qianxu(root,result);
        return result;
    }
    //前序遍历
    public void qianxu(TreeNode root,TreeNode result) {
        if(root.left != null){
            result.right = new TreeNode(root.left.val);
            qianxu(root.left,result.right);
        }
        if(root.right != null){
            result.left = new TreeNode(root.right.val);
            qianxu(root.right,result.left);
        }
    }
}

文章作者: 冯廷鑫
版权声明: 本博客所有文章除特別声明外,均采用 CC BY 4.0 许可协议。转载请注明来源 冯廷鑫 !
  目录