114 Flatten Binary Tree to Linked List – Medium

Problem:

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

Thoughts:

Idea is to use a modified preorder walk of the tree. Root, left, right.

Have a global pointer that keeps the node that needs a new right child to insert.

Solutions:

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
    TreeNode pointer = new TreeNode(-1);
    public void flatten(TreeNode root) {
        flattenTree(root);
    }
    private void flattenTree(TreeNode node){
        if (node != null){
            TreeNode left = node.left;
            TreeNode right = node.right;
            node.left = null;
            node.right = null;
            pointer.right = node;
            pointer = node;
            flattenTree(left);
            flattenTree(right);
        }
    }
}

Non-Recursion version:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void flatten(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = new TreeNode(-1);
        TreeNode node = root;
        while (true) {
            if (node == null) {
                if (stack.size() == 0) {
                    break;
                }
                node = stack.pop();
            }
            p.right = node;
            p = node;
            if (node.right != null) {
                stack.push(node.right);
            }
            node = node.left;
            p.left = null;
        }
    }
}