🌕 剑指 Offer 28. 对称的二叉树

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(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        // 思路：
        // 递归
        if(root == null) {
            return true;
        }
        return mySol(root.left, root.right);
    }

    private boolean mySol(TreeNode a, TreeNode b) {
        // 递归终止条件
        if(a == null && b == null) {
            return true;
        }
        if(a == null || b == null) {
            return false;
        }
        if(a.val != b.val) {
            return false;
        }
        return mySol(a.left, b.right) && mySol(a.right, b.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(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        // 思路：
        // 迭代，借助辅助栈，一次判断 4 个节点
        if(root == null) {
            return true;
        }
        LinkedList<TreeNode> stack = new LinkedList<>();
        stack.push(root.left);
        stack.push(root.right);
        while(!stack.isEmpty()) {
            TreeNode a = stack.pop();
            TreeNode b = stack.pop();
            if(a == null && b == null) {
                continue;
            }
            if(a == null || b == null) {
                return false;
            }
            if(a.val != b.val) {
                return false;
            }
            stack.push(a.left);
            stack.push(b.right);
            stack.push(a.right);
            stack.push(b.left);
        }
        return true;
    }
}

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输出 2