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;
}
}
}