Leetcode: Binary Tree Preorder Traversal

这道题一共有三道解法：

其中第二种解法，我参考了迭代器的写法，写了一个迭代器的类。重载了前++和—>操作符，但是这个解法会造成内存的超限，很奇怪，

解法1：递归解法：

 1 class Solution {
 2 public:
 3     vector<int> result;
 4     vector<int> preorderTraversal(TreeNode *root) {
 5 
 6         rpreorde(root);
 7         return result;
 8     }
 9     void rpreorde(TreeNode *root)
10     {
11          if(root == NULL)
12          return ;
13          result.push_back(root->val);
14          rpreorde(root->left);
15          rpreorde(root->right);
16 
17 
18     }
19 };

第二种解法迭代器

 1 class Solution {
 2 public:
 3     
 4     typedef TreeNode * STreeNode;
 5     vector<int> preorderTraversal(TreeNode *root) {
 6     
 7         stack<STreeNode> buf;
 8         nodeiterator temp(root,buf);
 9         vector<int> result;
10         
11         for(;temp.NotDone();++temp)
12         {
13           result.push_back(temp->val);
14 
15         }
16 
17     
18         return result;
19     }
20     class nodeiterator
21     {
22          public:
23 
24         stack<STreeNode>& buf;
25         STreeNode pcurrent;
26         nodeiterator(TreeNode *root,stack<STreeNode>& data):pcurrent(root),buf(data){}
27         void operator++()
28         {
29            if(pcurrent->left != NULL)
30            {
31                buf.push(pcurrent);
32                pcurrent = pcurrent->left;
33                return ;
34 
35            }
36            if(pcurrent->right != NULL)
37            {
38                pcurrent = pcurrent->right;
39                return ;
40            }
41            pcurrent = NULL;
42            while(buf.size()!=0)
43            {
44                pcurrent = buf.top();
45                buf.pop();
46                if(pcurrent->right != NULL)
47                {
48                   pcurrent = pcurrent->right;
49                      break;
50                }
51               
52                pcurrent = NULL;
53            }
54            
55         }
56         
57         TreeNode* operator->()
58         {
59              return pcurrent;
60             
61         }
62         TreeNode& operator*()
63         {
64              return *pcurrent;
65             
66         }
67         bool NotDone()
68         {
69             if(pcurrent == NULL)
70             return false;
71             else
72             return true;
73         }
74 
75     };
76 };

第三种解法：

class Solution {
public:
	
	typedef TreeNode * STreeNode;
    vector<int> preorderTraversal(TreeNode *root) {
	
		stack<STreeNode> buf;
	
		vector<int> result;
		STreeNode pcurrent;
		pcurrent = root;
		while(pcurrent != NULL)
		{
		   result.push_back(pcurrent->val);
		   if(pcurrent->left != NULL)
		   {
			   buf.push(pcurrent);
			   pcurrent = pcurrent->left;
			   continue;

		   }
		   if(pcurrent->right != NULL)
		   {
			   pcurrent = pcurrent->right;
			   continue ;
		   }
		   pcurrent = NULL;
		   while(buf.size()!=0)
		   {
			   pcurrent = buf.top();
			   buf.pop();
			   if(pcurrent->right != NULL)
			   {
			      pcurrent = pcurrent->right;
               	  break;
			   }
			  
			   pcurrent = NULL;
		   }
		   

		}
		

	
		return result;
    }
	
};

posted on 2014-12-20 12:00 xgcode 阅读(129) 评论(0) 收藏举报