README.md

101. 对称二叉树

English Version

题目描述

给定一个二叉树，检查它是否是镜像对称的。

 

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

 

进阶：

你可以运用递归和迭代两种方法解决这个问题吗？

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:
        if root is None:
            return True
        return self.is_symmetric(root.left, root.right)

    def is_symmetric(self, left: TreeNode, right: TreeNode) -> bool:
        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 self.is_symmetric(left.left, right.right) and self.is_symmetric(left.right, right.left)

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);
    }
}

C++

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        return isSymmetric(root->left, root->right);
    }

private:
    bool isSymmetric(TreeNode* left, TreeNode* right) {
        if (!left && !right) return true;
        if (!left && right || left && !right || 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 {number[][]}
 */
var levelOrder = function (root) {
  let result = [];
  if (!root) return result;

  let queue = [];
  queue.push(root);
  while (queue.length) {
    let size = queue.length;
    let levelItems = [];
    while (size--) {
      let node = queue.shift();
      levelItems.push(node.val);
      if (node.left) {
        queue.push(node.left);
      }
      if (node.right) {
        queue.push(node.right);
      }
    }
    result.push(levelItems);
  }
  return result;
};

### **...**

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