LeetCode algorithm binary tree—226. Flip binary tree

Table of contents

226. Flip a binary tree

Code:

operation result: 

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You are given the root node of a binary tree  root , flip the binary tree and return its root node.

Example 1:

Input: root = [4,2,7,1,3,6,9]
 Output: [4,7,2,9,6,3,1]

Example 2:

Input: root = [2,1,3]
 Output: [2,3,1]

Example 3:

Input: root = []
 Output: []

hint:

• The number of nodes in the tree [0, 100] is within
• -100 <= Node.val <= 100

Code:

/**
 * 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 { 
    // 从上到下左右子树的位置跟输入正好是相反的，递归交换
    public TreeNode invertTree(TreeNode root) {
        // 终止条件：当前节点为 null 时返回
        if(root==null) return null;
        // 交换当前节点的左右节点
        TreeNode tmp=root.right;
        root.right=root.left;
        root.left=tmp;
        // 递归的交换当前节点的左节点，右节点
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }

}

operation result: