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;
    }
};