Flatten Binary Tree to Linked List问题及解法

问题描述：

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

示例：
Given

         1        / \       2   5      / \   \     3   4   6

The flattened tree should look like:

   1    \     2      \       3        \         4          \           5            \             6

问题分析：

由题意可知，我们可以先把树的结点按照先序遍历存到数组vec中，然后遍历数组，修改指针，首尾相接即可。

过程详见代码：

/** * 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 preorder(TreeNode* root, vector<TreeNode*>& vec){        if(root == NULL)             return;        vec.push_back(root);        preorder(root->left, vec);        preorder(root->right, vec);    }            void flatten(TreeNode* root) {        vector<TreeNode*> vec;        preorder(root, vec);                for(int i = 0; i < (int)vec.size()-1; ++i){            vec[i]->right = vec[i+1];            vec[i]->left = NULL;        }                    }};