114. Flatten Binary Tree to Linked List

问题描述：

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / 
  2   5
 /    
3   4   6

The flattened tree should look like:

1
 
  2
   
    3
     
      4
       
        5
         
          6

解题思路：

因为要把所有的节点都放在右节点上。

我们可以用栈存储当前遇到的右节点，并将左节点放到右节点上并将左节点设为空（一开始我忘了。。TAT）！

当当前节点cur为空时，我们从栈中弹出一个节点并将其连接到上一个节点pre的右边。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
        if(!root)
            return;
        stack<TreeNode*> stk;
        TreeNode *cur = root;
        TreeNode *pre = root;
        while(cur || !stk.empty()){
            if(cur){
                if(cur->right)
                    stk.push(cur->right);
                cur->right = cur->left;
                cur->left = NULL;
                pre = cur;
                cur = cur->right;
            }else{
                cur = stk.top();
                pre->right = cur;
                stk.pop();
            }
        }
    }
};