Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

1   / \  2   5 / \   \3   4   6

The flattened tree should look like:

1 \  2   \    3     \      4       \        5         \          6

 

每个节点的右节点都是preorder traverse的下一个节点 -> 应该先 反preorder遍历 (即node.right->node.left->node) 到最后，从最后开始relink

time: O(n), space: O(height)

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    TreeNode prev = null;        public void flatten(TreeNode root) {        if(root == null) {            return;        }                flatten(root.right);        flatten(root.left);                root.right = prev;        root.left = null;        prev = root;    }}