【Leetcode長征系列】Flatten Binary Tree to Linked List

原題：

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

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

The flattened tree should look like:

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

思路：第一反應是深度優先遍歷，依次把節點放到樹葉子的右子樹位置即可。但是看到hint裏提到的是先序遍歷，就百度了下差別，現在概念更加清晰啦～

【先序，後序，中序針對二叉樹。深度、廣度針對普通樹。】

代碼如下：

/**
 * Definition for binary tree
 * 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;
        TreeNode *tmp;
        queue<TreeNode *> res;
        //depth first search
        stack<TreeNode*> tree;
        tree.push(root);
        while(!tree.empty()){
            tmp = tree.top();
            tree.pop();
            res.push(tmp);
            
            if(tmp->right!=NULL) tree.push(tmp->right);
            if(tmp->left!=NULL) tree.push(tmp->left);
        }
        
        root = res.front();
        res.pop();
        tmp = root;
        while(!res.empty()){
            tmp->right = res.front();
            tmp->left = NULL;
            tmp = tmp->right;
            res.pop();
        }
    }
};