Flatten Binary Tree to Linked List

32% Accepted

Flatten a binary tree to a fake "linked list" in pre-order traversal.

Here we use the right pointer in TreeNode as the next pointer in ListNode.

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Example
              1
               \
     1          2
    / \          \
   2   5    =>    3
  / \   \          \
 3   4   6          4
                     \
                      5
                       \
                        6
Note
Don't forget to mark the left child of each node to null.
Or you will get Time Limit Exceeded or Memory Limit Exceeded.

Challenge

Do it in-place without any extra memory.

Tags Expand

• Binary Tree
• Depth First Search

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void flatten(TreeNode root) {
        // write your code here
        if (root == null || (root.left == null && root.right == null)) {
            return;
        }

        TreeNode left = root.left;
        flatten(left);
        TreeNode right = root.right;
        flatten(right);

        root.left = null;
        root.right = left;
        while (root.right != null) {
            root = root.right;
        }
        root.right = right;

    }

}