LeetCode:Binary Tree Preorder Traversal(二叉树的先序遍历)

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

解法一：用栈实现（递归本质）

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> preorderTraversal(TreeNode* root) {
13         vector<int> result;
14         stack<TreeNode *> node_stack;
15         if(root!=NULL)
16             node_stack.push(root);
17         TreeNode *node;
18         
19         while(!node_stack.empty())
20         {   
21             node=node_stack.top();
22             result.push_back(node->val);
23             node_stack.pop();
24             
25             if(node->right!=NULL) node_stack.push(node->right);
26             if(node->left!=NULL) node_stack.push(node->left);
27         }
28         
29         return result;       
30         
31     }
32 };

 解法二：递归方式

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> preorderTraversal(TreeNode* root) {
13         vector<int> result;
14         preorder(root,result);
15         return result;
16         
17         
18     }
19     void preorder(TreeNode *node,vector<int> &result)
20     {   
21         if(node!=NULL)
22         {
23             result.push_back(node->val);
24             preorder(node->left,result);
25             preorder(node->right,result);
26         }
27         
28     }
29 };

 解法三：morris遍历方式 有待更新。

posted @ 2015-08-12 11:12  尾巴草  阅读(216)  评论(0)  编辑  收藏  举报