101. Symmetric Tree

最新推荐文章于 2025-09-05 16:34:31 发布

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最新推荐文章于 2025-09-05 16:34:31 发布

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CC 4.0 BY-SA版权

文章标签： java 数据结构与算法 runtime

原文链接：https://segmentfault.com/a/1190000018098639

本文介绍了一种检查二叉树是否为镜像的方法，即检查其是否围绕中心对称。通过递归方式对比左右子树节点，实现对二叉树对称性的判断。示例展示了对称与非对称二叉树的区别，并提供了代码实现。

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3
But the following [1,2,2,null,3,null,3] is not:
    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

难度：easy

题目：给定二叉树检查它是否为镜像。

思路：同same Tree

Runtime: 7 ms, faster than 46.76% of Java online submissions for Symmetric Tree.
Memory Usage: 30.1 MB, less than 0.94% of Java online submissions for Symmetric Tree.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (null == root) {
            return true;
        }
        return isSymmetric(root.left, root.right);
    }
    
    public boolean isSymmetric(TreeNode p, TreeNode q) {
        if (p == null && q != null || p != null && q == null) {
            return false;
        }
        if (p == null && q == null) {
            return true;
        }
        if (p.val != q.val) {
            return false;
        }
        
        return isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left);
    }
}