leetcode hot 05

合集 - leetcode hot100(25)

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收起

解题思路：前序遍历，然后将每个非空节点的左右子树交换（java要创建tmp变量交换），最后返回交换后的root；或者也可以创建一个新的二叉树节点然后赋值（比较麻烦）

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        treeReverse(root);
        return root;
    }

    public void treeReverse(TreeNode root) {
        if (root == null) return;
        
        // 交换左右子节点
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        
        // 递归反转左右子树
        treeReverse(root.left);
        treeReverse(root.right);
    }
}

/**
 * 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) {
        return treeReverse(root);
    }
    public TreeNode treeReverse(TreeNode root){
        if(root==null) return null;
        TreeNode root_back = new TreeNode(root.val);
        root_back.left = treeReverse(root.right);
        root_back.right = treeReverse(root.left);
        return root_back;
    }
}