[LeetCode]Binary Tree Preorder Traversal

Binary Tree Preorder Traversal

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].

Note: Recursive solution is trivial, could you do it iteratively?

 

不算难，使用stack即可。

 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         TreeNode* p;
15         stack<TreeNode*> mystack;
16         p = root;
17         if(p) mystack.push(p);
18         while(!mystack.empty())
19         {
20             p = mystack.top();
21             mystack.pop();
22             result.push_back(p->val);
23             if(p->right) mystack.push(p->right);
24             if(p->left) mystack.push(p->left);
25         }
26         return result;
27     }
28 };

 递归的方法很统一。先序，中序，后续，只要修改遍历的顺序即可。

 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         traversal(root,result);
15         return result;
16     }
17     void traversal(TreeNode* root,vector<int>& ret)
18     {
19         if(root)
20         {
21             ret.push_back(root->val);
22             traversal(root->left,ret);
23             traversal(root->right,ret);
24         }
25     }
26 };

 

posted @ 2015-09-04 22:45  Sean_le  阅读(133)  评论(0)    收藏  举报