[LeetCode Road] Same Tree - Solution/C++

100. Same Tree

Question:

Given the roots of two binary trees p and q, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical, and the nodes have the same value. ## Example:

Input: p = [1,2,3], q = [1,2,3]
Output: true

Input: p = [1,2], q = [1,null,2]
Output: false

Input: p = [1,2,1], q = [1,1,2]
Output: false

Constraints:

TThe number of nodes in both trees is in the range [0, 100].

-10^4 <= Node.val <= 10^4

Source code

Version 1

Idea:
This is my first time to process binary tree. After realizing question and example, it lets me recollect the solution of LeetCode 21.

I can use a similar if statement to code. But, I still don't know how to compare two binary trees of their child. So, I refered to programcreek.
Finally, I saw source code and understood the solution immediately. We can use recursion to bind the child checking (left & right).

Time complexity: O(n)
Space complexity: O(n)

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