1. 问题描述
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.
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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) {}
* };
*/
2. 解题思路
3. 代码
1 class Solution { 2 public: 3 bool isSameTree(TreeNode* p, TreeNode* q) 4 { 5 if (NULL == p && NULL == q) 6 { 7 return true; 8 } 9 10 if (NULL != p && NULL != q) 11 { 12 if (p->val == q->val) 13 { 14 return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); 15 } 16 else 17 { 18 return false; 19 } 20 } 21 return false; 22 } 23 #pragma region 非递归方式 24 bool isSameTree_1(TreeNode *p, TreeNode *q) 25 { 26 if(!isSameNode(p, q)) 27 return false; 28 if(!p && !q) 29 return true; 30 31 queue<TreeNode*> lqueue; 32 queue<TreeNode*> rqueue; 33 lqueue.push(p); 34 rqueue.push(q); 35 while(!lqueue.empty() && !rqueue.empty()) 36 { 37 TreeNode* lfront = lqueue.front(); 38 TreeNode* rfront = rqueue.front(); 39 40 lqueue.pop(); 41 rqueue.pop(); 42 43 if(!isSameNode(lfront->left, rfront->left)) 44 return false; 45 if(lfront->left && rfront->left) 46 { 47 lqueue.push(lfront->left); 48 rqueue.push(rfront->left); 49 } 50 51 if(!isSameNode(lfront->right, rfront->right)) 52 return false; 53 if(lfront->right && rfront->right) 54 { 55 lqueue.push(lfront->right); 56 rqueue.push(rfront->right); 57 } 58 } 59 return true; 60 } 61 bool isSameNode(TreeNode* p, TreeNode *q) 62 { 63 if(!p && !q) 64 return true; 65 if((p && !q) || (!p && q) || (p->val != q->val)) 66 return false; 67 return true; 68 } 69 #pragma endregion 70 };