17
05/2015
[LeetCode] Same Tree
Same Tree
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.
解题思路:
递归做即可。两颗二叉树相同,当且仅当当前节点值相同,并且对应的左右孩子树也相同。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSameTree(TreeNode* p, TreeNode* q) { if(p==NULL && q==NULL){ return true; }else if(p==NULL || q==NULL){ //其中有一个是空 return false; } if(p->val!=q->val){ return false; } return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); } };
另外一种非递归法,用队列来记录两棵树当前状态:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSameTree(TreeNode* p, TreeNode* q) { if(p==NULL && q==NULL){ return true; }else if(p==NULL || q==NULL){ //其中有一个是空 return false; } //逻辑上,两个队列的大小相同 queue<TreeNode*> queue1({p}); queue<TreeNode*> queue2({q}); while(!queue1.empty()){ TreeNode* node1 = queue1.front(); TreeNode* node2 = queue2.front(); queue1.pop(); queue2.pop(); if(node1->val!=node2->val){ return false; } if(node1->left!=NULL && node2->left!=NULL){ queue1.push(node1->left); queue2.push(node2->left); }else if(node1->left!=NULL || node2->left!=NULL){ return false; } if(node1->right!=NULL && node2->right!=NULL){ queue1.push(node1->right); queue2.push(node2->right); }else if(node1->right!=NULL || node2->right!=NULL){ return false; } } return true; } };
二次刷题(2015-08-03)
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSameTree(TreeNode* p, TreeNode* q) { if(p==NULL && q==NULL){ return true; } if(p==NULL || q==NULL){ return false; } return p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right); } };
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