100. Same Tree

100. Same Tree

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Total Accepted: 100129  Total Submissions: 236623  Difficulty: Easy

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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朴素的递归思想：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //思路首先：只要p和q都不为null且val相等就继续判断p的左子树和q的左子树以及p的右子树和q的右子树
 //如果出现某子树存在而另一颗树的相应子树却不存在，则返回false
class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if(p==NULL&&q==NULL)  
            return true;  
        else if(p==NULL&&q!=NULL)  
            return false;  
        else if(p!=NULL&&q==NULL)  
            return false;  
        else if(p!=NULL&&q!=NULL && p->val!=q->val)  
            return false;  
        else  
            return (isSameTree(p->left,q->left))&&(isSameTree(p->right,q->right));  
    }
};

广度优先迭代遍历是否一样：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //思路首先：递归能做，显然迭代也能做，下面采用了广度优先遍历两棵树是否一样
class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
      if(q==NULL && p ==NULL)  
        return true; 
      if(q == NULL && p!=NULL || q != NULL && p==NULL)    
            return false;
      //广度优先遍历两个子树是否一样  
      queue<TreeNode*> que1,que2;    
      TreeNode* curNode1=p;
      TreeNode* curNode2=q;      
      que1.push(curNode1);  
      que2.push(curNode2);
      while (!que1.empty()&&!que2.empty())    
      {      
         curNode1=que1.front();//出队首元素      
         que1.pop();//删除队首元素  
         curNode2=que2.front();//出队首元素      
         que2.pop();//删除队首元
         if(!(curNode1->val==curNode2->val))
            return false;
         if(curNode1->left!=NULL && curNode2->left!=NULL)    
         {   
            que1.push(curNode1->left);    
            que2.push(curNode2->left); 
         }
         
         if(curNode1->right!=NULL && curNode2->right!=NULL)    
         {     
             que1.push(curNode1->right);
             que2.push(curNode2->right); 
         }
         if(curNode1->left == NULL && curNode2->left!=NULL || curNode1->left != NULL && curNode2->left==NULL)    
            return false;
            
         if(curNode1->right == NULL && curNode2->right!=NULL || curNode1->right != NULL && curNode2->right==NULL)    
            return false;
      }  
      return true;  
    }
};

前序式深度优先搜索来做：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //思路首先：递归能做，显然迭代也能做，下面采用了前序式深度优先遍历两棵树是否一样
class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
      if(q==NULL && p ==NULL)  
        return true; 
      if(q == NULL && p!=NULL || q != NULL && p==NULL)    
            return false;
      //前序式深度优先遍历两个子树是否一样  
      stack<TreeNode*> stk1,stk2;    
      TreeNode* curNode1=p;
      TreeNode* curNode2=q;      
      stk1.push(curNode1);  
      stk2.push(curNode2);
      while (!stk1.empty()&&!stk2.empty())    
      {      
         curNode1=stk1.top();//出队首元素      
         stk1.pop();//删除队首元素  
         curNode2=stk2.top();//出队首元素      
         stk2.pop();//删除队首元
         if(!(curNode1->val==curNode2->val))
            return false;
         if(curNode1->left!=NULL && curNode2->left!=NULL)    
         {   
            stk1.push(curNode1->left);    
            stk2.push(curNode2->left); 
         }
         
         if(curNode1->right!=NULL && curNode2->right!=NULL)    
         {     
             stk1.push(curNode1->right);
             stk2.push(curNode2->right); 
         }
         if(curNode1->left == NULL && curNode2->left!=NULL || curNode1->left != NULL && curNode2->left==NULL)    
            return false;
            
         if(curNode1->right == NULL && curNode2->right!=NULL || curNode1->right != NULL && curNode2->right==NULL)    
            return false;
      }  
      return true;  
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50507131

原作者博客：http://blog.csdn.net/ebowtang