题目：

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

Hints:

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

思路：

递归，先flatten左边的，再flatten右边的，然后进行连接

package tree;public class FlattenBinaryTreeToLinkedList {    public void flatten(TreeNode root) {        if (root == null) return;        flatten(root.left);        flatten(root.right);        if (root.left != null) {            TreeNode tmp = root.right;            root.right = root.left;            root.left = null;            TreeNode rightMostNode = root;            while (rightMostNode.right != null) {                rightMostNode = rightMostNode.right;            }            rightMostNode.right = tmp;        }    }}