🌗 226. 翻转二叉树

难度: 🌗

问题描述

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解法 1 - 递归

/**
 * 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) {
        // 思路：
        // 递归
        return mySol(root);
    }

    private TreeNode mySol(TreeNode root) {
        // 递归终止条件
        if(root == null) {
            return root;
        }
        // root != null
        TreeNode left = mySol(root.left);
        TreeNode right = mySol(root.right);
        // 交换左右节点
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
        return root;
    }
}

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输出 1

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解法 2 - 迭代

/**
 * 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) {
        // 思路：
        // 迭代 - 借助辅助栈
        LinkedList<TreeNode> stack = new LinkedList<>();
        if(root == null) {
            return root;
        }
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode cur = stack.pop(); // 要翻转的二叉树节点
            swap(cur);
            if(cur.left != null) {
                stack.push(cur.left);
            }
            if(cur.right != null) {
                stack.push(cur.right);
            }
        }
        return root;
    }

    private void swap(TreeNode root) {
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
    }
}

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输出 2