[leetcode] Binary Tree Inorder Traversal

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].

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

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}".

使用非递归的方法中序遍历二叉树。

使用一个栈来模拟递归过程，如果游走指针不为NULL，就把节点入栈，继续访问其左孩子。否则的话访问栈顶元素，并把指针指向他的右孩子。访问栈顶和跳转到右孩子一定要在一个语句块中实现，原因有二，其一为从栈顶弹出的指针的左孩子一定已经访问过了，所以按中序遍历的顺序应该访问该节点，其二这样做就无需判断游走指针指向的节点的左孩子是否访问过，应为栈顶的元素的左孩子一定访问过，只需访问其右孩子即可。

代码如下：

 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 

11 class Solution {

12     public:

13         vector<int> inorderTraversal(TreeNode *root) {

14             vector<int> res;

15             if( root == NULL )

16             {

17                 return res;

18             }

19             TreeNode * tr = root;

20 

21             stack<TreeNode*> s;

22             while(s.size()>0 || tr != NULL)

23             {

24                 if(tr!=NULL)

25                 {

26                     s.push(tr);

27                     tr= tr->left;

28                 }

29                 else

30                 {

31                     tr = s.top();

32                     s.pop();

33                     res.push_back(tr->val);

34                     tr = tr->right;

35                 }

36             }

37             return res;

38         }

39 };

并附上其递归版本：

 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     void T(TreeNode * root,vector<int> & r)

13     {

14         if(root==NULL) return;

15         T(root->left,r);

16         r.push_back(root->val);

17         T(root->right,r);

18     }

19     vector<int> inorderTraversal(TreeNode *root) {

20         // Start typing your C/C++ solution below

21         // DO NOT write int main() function

22         vector<int> r;

23         T(root,r);

24         return r;

25     }

26 };