# Flatten Binary Tree to Linked List

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.

## Solution

Recursion. Return the last element of the flattened sub-tree.

## Code

``````public class Solution {
public void flatten(TreeNode root) {
flatten_3(root);
}

public void flatten_1(TreeNode root) {
if (root == null) return;
flatten_1(root.left);
flatten_1(root.right);
if (root.left == null) return;
TreeNode node = root.left;
while (node.right != null) node = node.right;
node.right = root.right;
root.right = root.left;
root.left = null;
}

public void flatten_2(TreeNode root) {
if (root == null) return;
Stack<TreeNode> stk = new Stack<TreeNode>();
stk.push(root);
while (stk.empty() == false) {
TreeNode cur = stk.pop();
if (cur.right != null) stk.push(cur.right);
if (cur.left != null) stk.push(cur.left);
cur.left = null;
cur.right = stk.empty() == true ? null : stk.peek();
}
}

public TreeNode flattenRe3(TreeNode root, TreeNode tail) {
if (root == null) return tail;
root.right = flattenRe3(root.left, flattenRe3(root.right, tail));
root.left = null;
return root;
}
public void flatten_3(TreeNode root) {
if (root == null) return;
flattenRe3(root, null);
}
}
``````