Cinte's Leetcode Record
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100. Same Tree

Posted on Edited on In Disqus:
Symbols count in article: 1.7k Reading time ≈ 2 mins.

100. Same Tree

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/**
 * 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) {}
 * };
 */
/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {
        if((p && !q) || (!p && q))
        {
            return false;
        }else if(!p && !p)
        {
            return true;
        }else if(p->val == q->val)
        {
            return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
        }
        return false;
    }
};

T(n) : O(N) S(n) : O(logN) 当平衡时，O(N) 当不平衡时，空间为递归所用的stack的大小

review
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/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {
        if(!p)
            return !q;
        if(!q)
            return !p;
        return p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};