🌗101. 对称二叉树

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 boolean isSymmetric(TreeNode root) {
        // 思路:
        // 递归
        // 不用想怎么写，写到哪里自然就写出来了
        return mySol(root.left, root.right);
    }

    private boolean mySol(TreeNode left, TreeNode right) {
        // 递归终止条件
        if(left == null && right == null) {
            return true;
        }
        if(left == null || right == null) {
            return false;
        }
        // left != null && right != null
        if(left.val != right.val) {
            return false;
        }
        // left.val == right.val
        return mySol(left.left, right.right) && mySol(left.right, right.left);
    }
}

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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 boolean isSymmetric(TreeNode root) {
        // 思路：
        // 迭代 - 借助 辅助栈
        LinkedList<TreeNode> stack = new LinkedList<>(); // 一次同时压左右节点
        stack.push(root.left);
        stack.push(root.right);
        while(!stack.isEmpty()) {
            TreeNode right = stack.pop();
            TreeNode left = stack.pop();
            // 逻辑判断
            if(left == null && right == null) {
                continue;
            } else if(left == null || right == null) {
                return false;
            } else if(left.val != right.val) {
                return false;
            } else {
                stack.push(left.left);
                stack.push(right.right);
                stack.push(left.right);
                stack.push(right.left);
            }
        }
        return true;
    }
}

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输出 2