[LeetCode] 101. Symmetric Tree

[frdmu]

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

 

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -100 <= Node.val <= 100
   

Follow up: Could you solve it both recursively and iteratively?
 
 

解法一：
递归。比较左右子树。代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool check(TreeNode* left, TreeNode* right) {
        if (left && !right) return false;
        if (right && !left) return false;
        if (!left && !right) return true;

        if (left->val == right->val) {
            return check(left->left, right->right) && check(left->right, right->left);
        }
        return false;
    }
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;

        return check(root->left, root->right);
    }
};

解法二：
迭代法。根据官方题解，一次队列里放两个对应位置的结点。代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        
        queue<TreeNode*> q;
        q.push(root->left);
        q.push(root->right);
        while (!q.empty()) {
            TreeNode* left = q.front();
            q.pop();
            TreeNode* right = q.front();
            q.pop();
            
            if (!left && !right) continue; 
            if (!left || !right || (left->val != right->val)) {
                return false;
            }
            
            q.push(left->left);
            q.push(right->right);
            q.push(left->right);
            q.push(right->left);
        }
        
        return true;
    }
};