(M)DFS:114. Flatten Binary Tree to Linked List

这道题的意思是，把一个二叉树展开成一个链。

参考了大神的做法：

把左子树展开成链，把右子树展开成链，然后让右子树变成展开后的左子树，遍历到树的最下端，把展开成的右子树加在这个最下端。

/**
 * 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;
        if(root->left)  flatten(root->left);
        if(root->right) flatten(root->right);
        TreeNode *tmp = root->right;
        root->right = root->left;
        root->left = NULL;
        while(root->right != NULL)
            root = root->right;
        root->right = tmp;
    }
};