LeetCode No.114 Flatten Binary Tree to Linked List

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

====================================================================================
题目链接：https://leetcode.com/problems/flatten-binary-tree-to-linked-list/

题目大意：将二叉树按照先序遍历的顺序转换成一个只有右子树的二叉树（一条链表），要求空间复杂度为O(1)。

思路1：通过先序遍历边搜边转换，由于过程的操作有点复杂，我选择用另外一种做法。

思路2：递归调用，假设函数flattenTree能符合题意地转成一棵新的二叉树树，则root节点的右子树等于flattenTree（root的左子树），转换完之后左子树的最后一个节点的右子树为等于flattenTree（root的右子树），同时root的左子树为NULL。

附上代码：

/** * 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 == NULL )            return ;        root = flattenTree ( root ) ;    }    TreeNode* flattenTree ( TreeNode* root )    {        if ( root == NULL )            return root ;        if ( root -> left == NULL )        {            root -> right = flattenTree ( root -> right ) ;            return root ;        }                    TreeNode* temp = root -> right ;        root -> right = flattenTree ( root -> left ) ;        TreeNode* t = root ;        while ( t -> right != NULL )            t = t -> right ;        t -> right = flattenTree ( temp ) ;        root -> left = NULL ;        return root ;    }};