LeetCode(100)题解--Same Tree

https://leetcode.com/problems/same-tree/

题目：

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.

思路： DFS

AC代码：

1.递归

/**
* 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){
bool ju;
if(p->val==q->val)
ju=true;
else
return false;
if (p->left==NULL && q->left==NULL)
;
else if(p->left!=NULL && q->left!=NULL)
ju=ju&&isSameTree(p->left,q->left);
else
return false;
if (p->right==NULL && q->right==NULL)
;
else if(p->right!=NULL && q->right!=NULL)
ju=ju&&isSameTree(p->right,q->right);
else
return false;
return ju;
}
else
return false;
}
};

2.非递归

/**
* 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){
stack<TreeNode*> ms1,ms2;
ms1.push(p);
ms2.push(q);
TreeNode* p1,*q1;
while(!ms1.empty()){
p1=ms1.top();
q1=ms2.top();
if(p1->val!=q1->val)
return false;
else{
ms1.pop();
ms2.pop();
if(p1->right==NULL&&q1->right==NULL)
;
else if(p1->right!=NULL&&q1->right!=NULL){
ms1.push(p1->right);
ms2.push(q1->right);
}
else
return false;
if(p1->left==NULL&&q1->left==NULL)
;
else if(p1->left!=NULL&&q1->left!=NULL){
ms1.push(p1->left);
ms2.push(q1->left);
}
else
return false;
}
}
return true;
}
else
return false;
}
};