# Binary Tree Inorder/Preorder Traversal 返回中序和前序/遍历二叉树的元素集合

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

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

   1
\
2
/
3


return [1,3,2].

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
/ \
2   3
/
4
\
5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

 1 /**
2  * Definition for binary tree
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> inorderTraversal(TreeNode *root) {
13         vector<int> ret;
14         if(root == NULL)
15             return ret;
16
17         stack<TreeNode *> stack;
18         stack.push(root);
19
20         while(!stack.empty()){
21             TreeNode *node = stack.top();
22             stack.pop();
23             if(node->left == NULL && node->right == NULL){
24                 ret.push_back(node->val);
25             }
26             else{
27                 if(node->right != NULL)
28                     stack.push(node->right);
29                 stack.push(node);
30                 if(node->left != NULL)
31                     stack.push(node->left);
32
33                 node->left = node->right = NULL;
34             }
35
36         }
37
38         return ret;
39
40     }
41 };

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].

 1 /**
2  * Definition for binary tree
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> ret;
14         if(root == NULL)
15             return ret;
16         PreorderTraversal(root,ret);
17         return ret;
18     }
19
20     void PreorderTraversal(TreeNode *root,vector<int> &ret){
21         if(root != NULL){
22             ret.push_back(root->val);
23             PreorderTraversal(root->left,ret);
24             PreorderTraversal(root->right,ret);
25         }
26     }
27 };

posted @ 2014-10-21 21:44  非著名程序师  阅读(509)  评论(0编辑  收藏