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.
题目大意：
给定两个二叉树，写一个函数检查它们是否相等。两个二叉树如果结构相同并且对应的节点有相同的值，则被视作为相等。
递归解法：
1 /** 2 * Definition for a binary tree node. 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 //对应的节点同时为空，返回true 14 if(p == NULL && q == NULL) 15 return true; 16 //其他情况!p&&q，p&&!q, p&&q&&p->val != q->val 都返回false 17 else if(!p && q || p && !q || (p->val != q->val)) 18 return false; 19 return (isSameTree(p->left, q->left) && isSameTree(p->right, q->right)); 20 } 21 };
非递归：
建立两个队列分别进行层次遍历，进队时检查对应点是否相等
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 if(!isSameNode(p, q)) 14 return false; 15 if(!p && !q) 16 return true; 17 18 queue<TreeNode*> lqueue; 19 queue<TreeNode*> rqueue; 20 lqueue.push(p); 21 rqueue.push(q); 22 while(!lqueue.empty() && !rqueue.empty()) 23 { 24 TreeNode* lfront = lqueue.front(); 25 TreeNode* rfront = rqueue.front(); 26 27 lqueue.pop(); 28 rqueue.pop(); 29 30 if(!isSameNode(lfront->left, rfront->left)) 31 return false; 32 if(lfront->left && rfront->left) 33 { 34 lqueue.push(lfront->left); 35 rqueue.push(rfront->left); 36 } 37 38 if(!isSameNode(lfront->right, rfront->right)) 39 return false; 40 if(lfront->right && rfront->right) 41 { 42 lqueue.push(lfront->right); 43 rqueue.push(rfront->right); 44 } 45 } 46 return true; 47 } 48 bool isSameNode(TreeNode* p, TreeNode *q) 49 { 50 if(!p && !q) 51 return true; 52 if((p && !q) || (!p && q) || (p->val != q->val)) 53 return false; 54 return true; 55 } 56 };