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symmetric-tree

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Symmetric Tree – LeetCode 101

Problem

Description

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

Answer

Original

Code

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/**
 * 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* p, TreeNode* q){
        if(p != nullptr && q != nullptr){
            if(p->val == q->val)
                return issymmetric(p->right,q->left) && issymmetric(p->left,q->right);
            else 
                return false;
        } else if (p == nullptr && q == nullptr) return true;
            else {return false;}
    }
    
    bool isSymmetric(TreeNode* root) {
        if(!root) return true;
        return issymmetric(root->left,root->right);
    }
};

思路

复用上题的代码，把一棵树交换左右节点分解成两颗树即可。时间复杂度$O(n)$，$n$为节点数，空间复杂度$O(n)$。
耗时$6$ ms，排名$96.44\%$

Better

思路

同样使用DFS的迭代来实现，时间复杂度$O(n)$，$n$为节点数，空间复杂度$O(n)$。也可以看LeetCode官方解。
耗时$6$ ms，排名$96.44\%$

Code

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/**
 * 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) {
        queue<TreeNode *> q;
        q.push(root);
        q.push(root);
        while(!q.empty()){
            TreeNode *t1 = q.front(); q.pop();
            TreeNode *t2 = q.front(); q.pop();
            if(!t1 && !t2) continue;
            if(!t1 || !t2) return false;
            if(t1->val != t2->val) return false;
            q.push(t1->left);
            q.push(t2->right);
            q.push(t1->right);
            q.push(t2->left);
        }
        return true;
    }
};