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.
判断两个树是否相同，起是用递归很好解决，但是我一开始居然想使用前序加中序遍历来唯一确定一棵树：
1 class Solution { 2 public: 3 bool isSameTree(TreeNode* p, TreeNode* q) { 4 vector<int>res1; 5 vector<int>res2; 6 if(p == NULL && q == NULL) 7 return true; 8 else if(p == NULL || q == NULL) 9 return false; 10 preorderTraversal(p, res1); 11 preorderTraversal(q, res2); 12 bool ans1 = res1 == res2; 13 res1.clear(), res2.clear(); 14 infixTraversal(p, res1); 15 infixTraversal(q, res2); 16 bool ans2 = res1 == res2; 17 return res1 && res2; 18 19 } 20 void preorderTraversal(TreeNode * p, vector<int> & res) 21 { 22 if(p == NULL) return; 23 res.push_back(p->val); 24 preorderTraversal(p->left, res); 25 preorderTraversal(p->right, res); 26 } 27 void infixTraversal(TreeNode * p, vector<int> & res) 28 { 29 if(p == NULL) return; 30 infixTraversal(p->left, res); 31 res.push_back(p->val); 32 infixTraversal(p->right, res); 33 } 34 };
然而并不可行，只能使用下面这一种方法：
1 class Solution { 2 public: 3 bool isSameTree(TreeNode* p, TreeNode* q) { 4 if(!p && !q) 5 return true; 6 else if(!p && q) 7 return false; 8 else if(p && !q) 9 return false; 10 else{ 11 if(p->val != q->val) 12 return false; 13 else 14 return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); 15 } 16 } 17 };
当然也可以采用迭代的方式来完成：
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 queue<TreeNode *> pQueue; 14 queue<TreeNode *> qQueue; 15 pQueue.push(p); //这里只能用队列而不能用栈，因为单向链表的性质决定 16 qQueue.push(q); 17 while(!pQueue.empty() && !qQueue.empty()){ 18 TreeNode * pTmp = pQueue.front(); 19 TreeNode * qTmp = qQueue.front(); 20 pQueue.pop(), qQueue.pop(); 21 if(!pTmp && !qTmp) 22 continue; 23 else if(pTmp && !qTmp || !pTmp && qTmp || pTmp->val != qTmp->val) 24 return false; 25 else{ 26 pQueue.push(pTmp->left); 27 pQueue.push(pTmp->right); 28 qQueue.push(qTmp->left); 29 qQueue.push(qTmp->right); 30 } 31 } 32 return pQueue.empty() && qQueue.empty(); 33 } 34 };
下面是java代码：
递归的：
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public boolean isSameTree(TreeNode p, TreeNode q) { 12 if(p == null && q == null) 13 return true; 14 else if(p == null) 15 return false; 16 else if(q == null) 17 return false; 18 else{ 19 if(p.val == q.val){ 20 return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); 21 }else 22 return false; 23 } 24 } 25 }
迭代版：
1 public class Solution { 2 public boolean isSameTree(TreeNode p, TreeNode q) { 3 Queue<TreeNode> queueP = new LinkedList<TreeNode>(); 4 Queue<TreeNode> queueQ = new LinkedList<TreeNode>(); 5 TreeNode p1, p2; 6 queueP.add(p); 7 queueQ.add(q); 8 while(!queueP.isEmpty() && !queueQ.isEmpty()){ 9 p1 = queueP.poll(); 10 p2 = queueQ.poll(); 11 if(p1 == null && p2 != null || 12 p1 != null && p2 == null) 13 return false; 14 if(p1 == null && p2 == null) 15 continue; 16 if(p1.val == p2.val){ 17 queueP.add(p1.left); 18 queueP.add(p1.right); 19 queueQ.add(p2.left); 20 queueQ.add(p2.right); 21 }else return false; 22 } 23 if(queueP.isEmpty() && queueQ.isEmpty()) 24 return true; 25 return false; 26 } 27 }