LeetCode: 101. Symmetric Tree

LeetCode: 101. Symmetric Tree

题目描述

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.
题目大意: 判断给定二叉树是否是对称的。

解题思路

判断给定二叉树是否是对称的，只需要判断其左右子树是否对称。而判断两棵二叉树(T1，T2)是否对称，只需要判断:
1. 树根的元素是否相等?
2. T1 的左子树是否和 T2 的右子树对称？
3. T1 的右子树是否和 T2 的左子树对称？

AC代码

• 递归解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    // 判断两个二叉树是否对称
    bool isSymmetric(TreeNode* leftTree, TreeNode* rightTree)
    {
        if(leftTree == nullptr || rightTree == nullptr) return (leftTree == rightTree);
        if(leftTree->val != rightTree->val) return false;
        return isSymmetric(leftTree->left, rightTree->right) &&
         isSymmetric(leftTree->right, rightTree->left);
    }
public:
    bool isSymmetric(TreeNode* root) {
        if(root == nullptr) return true;
        return isSymmetric(root->left, root->right);
    }
};

• 迭代解法

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root == nullptr) return true;
        queue<TreeNode*> que; 
        bool isSym = true;
        que.push(root->left);
        que.push(root->right);

        while(!que.empty())
        {
            TreeNode* leftTree = que.front();
            que.pop();
            TreeNode* rightTree = que.front();
            que.pop();
            if(leftTree == nullptr || rightTree == nullptr) isSym = (leftTree == rightTree);
            else if(leftTree->val != rightTree->val) isSym = false;
            else
            {
                que.push(leftTree->left);
                que.push(rightTree->right);
                que.push(leftTree->right);
                que.push(rightTree->left);
            }

            if(isSym == false) break;
        }

        return isSym;
    }
};