2013.12.31 22:56
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
Solution:
Check the root, the left subtree and the right subtree, recursively. That's the answer.
Time complexity is O(min(m, n)), where m and n are the numbers of nodes in both trees.
Space complexity is O(1).
Accepted code:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSameTree(TreeNode *p, TreeNode *q) { 13 // Note: The Solution object is instantiated only once and is reused by each test case. 14 if(p == nullptr || q == nullptr){ 15 return p == q; 16 }else{ 17 if(p->val == q->val){ 18 return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); 19 }else{ 20 return false; 21 } 22 } 23 } 24 };