100.判断两个二叉树是否完全相等 Same Tree


递归实现
  1. static public bool IsSameTree(TreeNode root1, TreeNode root2) {
  2. 	if (root1 == null && root2 == null) {
  3. 		return true;
  4. 	}
  5. 	if ((root1 == null && root2 != null) || (root1 != null && root2 == null)) {
  6. 		return false;
  7. 	}
  8. 	if (root1.val != root2.val) {//判断每个节点的值是否相等，如果去除此判断，则判断两个二叉树是否结构相等
  9. 		return false;
  10. 	}
  11. 	return IsSameTree(root1.left, root2.left) && IsSameTree(root1.right, root2.right);
  12. }


非递归实现方式
  1. bool  BTreeCompare（BTreeNode_t *pRoot1, BTreeNode_t *pRoot2)  
  2. {  
  3.     if( pRoot1 == NULL && pRoot2 == NULL )  
  4.         return false;  
  5.   
  6.   
  7.     queue <BTreeNode_t *> que1;  
  8.     queue <BTreeNode_t *> que2;  
  9.   
  10.   
  11.     que1.push(pRoot1);  
  12.     que2.push(pRoot2);  
  13.   
  14.   
  15.     int curLevelNodeTotal1 = 0;  
  16.     int curLevelNodeTotal2 = 0;  
  17.   
  18.   
  19.     bool flag = true; //作为比较不一致时跳出标识  
  20.     while( ( !que1.empty()) && ( !que2.empty())) //当两个队列均不为空时，才进行比较  
  21.     {  
  22.         curLevelNodeTotal1 = que1.size();  //获取树1的当前层节点总数  
  23.         curLevelNodeTotal2 = que2.size(); //获取树2的当前层节点总数  
  24.         if( curLevelNodeTotal1 != curLevelNodeTotal2){  
  25.             flag = false;//当前层节点总数都不一致，不需要比较了，直接跳出  
  26.             break;  
  27.         }  
  28.         int cnt1 = 0;//遍历本层节点时的计数器  
  29.         int cnt2 = 0;  
  30.         while( cnt1 < curLevelNodeTotal1  && cnt2 < curLevelNodeTotal2){  
  31.             ++cnt1;  
  32.             ++cnt2;  
  33.             pRoot1 = que1.front();  
  34.             que1.pop();  
  35.             pRoot2 = que2.front();  
  36.             que2.pop();  
  37.   
  38.             //比较当前节点中数据是否一致  
  39.             if( pRoot1->m_pElemt != pRoot2->m_pElemt ){  
  40.                 flag = false;  
  41.                 break;  
  42.             }  
  43.             //判断pRoot1和pRoot2左右节点结构是否相同  
  44.             if( ( pRoot1->m_pLeft != NULL && pRoot2->m_pLeft == NULL )    ||  
  45.                 ( pRoot1->m_pLeft == NULL && pRoot2->m_pLeft != NULL )    ||  
  46.                 ( pRoot1->m_pRight != NULL && pRoot2->m_pRight == NULL )    ||  
  47.                 ( pRoot1->m_pRight == NULL && pRoot2->m_pRight != NULL )  
  48.             ){  
  49.                 flag = false;  
  50.                 break;  
  51.             }  
  52.    
  53.             //将左右节点入队  
  54.             if( pRoot1->m_pLeft != NULL )  
  55.                 que1.push( pRoot1->m_pLeft);  
  56.             if( pRoot1->m_pRight != NULL )  
  57.                 que1.push( pRoot1->m_pRight);  
  58.             if( pRoot2->m_pLeft != NULL )  
  59.                 que2.push( pRoot2->m_pLeft);  
  60.             if( pRoot2->m_pRight != NULL )  
  61.                 que2.push( pRoot2->m_pRight);  
  62.         }  
  63.   
  64.   
  65.         if( flag == false )  
  66.             break;  
  67.     }  
  68.   
  69.     //如果比较标志为false，则不相同  
  70.     if( flag == false ){  
  71.         while( !que1.empty() )  
  72.             que1.pop();  
  73.         while( !que2.empty())  
  74.             que2.pop();  
  75.   
  76.   
  77.         return false;  
  78.     }  
  79.   
  80.   
  81.     return true;  
  82. }





posted @ 2017-01-10 23:07  xiejunzhao  阅读(3026)  评论(0)    收藏  举报