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;
}
} }

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