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

click to show hints.

Hints:

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

解题思路：

题意为将二叉树依照先序遍历压平。

最開始时对in-place理解错误，以为空间复杂度仅仅能为O(1)。事实上不然（我们姑且觉得递归的时候空间不变）。

我们能够用递归来解决问题。

首先定义个递归函数，改递归函数返回root为根节点的最右节点。

倘若root->left不为空。则将左子树的节点插入到root->right中。

递归代码例如以下：

/**
 * 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) {
        flattenHelper(root);
    }
    
    //返回root为根的最右边的那个节点
    TreeNode* flattenHelper(TreeNode* root){
        if(root == NULL){
            return NULL;
        }
        TreeNode* leftMost = flattenHelper(root->left);
        TreeNode* rightMost = flattenHelper(root->right);
        if(leftMost!=NULL){
            TreeNode* temp = root->right;
            root->right = root->left;
            root->left = NULL;
            leftMost->right = temp;
        }
        
        if(leftMost==NULL&&rightMost==NULL){
            return root;
        }else if(rightMost==NULL){
            return leftMost;
        }else{
            return rightMost;
        }
    }
};