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面试题 28. 对称的二叉树

题目描述

请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样，那么它是对称的。

例如，二叉树  [1,2,2,3,4,4,3] 是对称的。

    1
   / \
  2   2
 / \ / \
3  4 4  3

但是下面这个  [1,2,2,null,3,null,3] 则不是镜像对称的:

    1
   / \
  2   2
   \   \
   3    3

示例 1：

输入：root = [1,2,2,3,4,4,3]
输出：true

示例 2：

输入：root = [1,2,2,null,3,null,3]
输出：false

限制：

• 0 <= 节点个数 <= 1000

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        def is_symmetric(left, right):
            if left is None and right is None:
                return True
            if left is None or right is None or left.val != right.val:
                return False
            return is_symmetric(left.left, right.right) and is_symmetric(left.right, right.left)
        if root is None:
            return True
        return is_symmetric(root.left, root.right)

Java

/**
 * 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) {
        if (root == null) return true;
        return isSymmetric(root.left, root.right);
    }

    private boolean isSymmetric(TreeNode left, TreeNode right) {
        if (left == null && right == null) return true;
        if (left == null || right == null || left.val != right.val) return false;
        return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function (root) {
  function dfs(left, right) {
    if (!left && !right) return true;
    if (!left || !right || left.val != right.val) return false;
    return dfs(left.left, right.right) && dfs(left.right, right.left);
  }
  if (!root) return true;
  return dfs(root.left, root.right);
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSymmetric(root *TreeNode) bool {
    if root == nil {
        return true
    }
    return isSymme(root.Left, root.Right)
}

func isSymme(left *TreeNode, right *TreeNode) bool {
    if left == nil && right == nil {
        return true
    }
    if left == nil || right == nil || left.Val != right.Val {
        return false
    }
    return isSymme(left.Left, right.Right) && isSymme(left.Right, right.Left)
}

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