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

解题思路：
1. 如果根节点为空，返回，此为递归终止条件。
2. 左右子树flatten
3. 根节点与flatten后的左右子树结合成linked list
3.1 如果左子树为空，无需操作，返回
3.2 找到左子树的最右节点，与右子树连接，根节点与左子树连接。

/** * 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 == nullptr) return;        flatten(root->left);        flatten(root->right);        if (nullptr == root->left) return;        TreeNode *p = root->left;        while (p->right) p = p->right;        p->right = root->right;        root->right = root->left;        root->left = nullptr;    }};