Flatten Binary Tree to Linked List

update Jul 14, 2017 14:43

Lintcode

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.

Notice

Don't forget to mark the left child of each node to null. Or you will get Time Limit Exceeded or Memory Limit Exceeded.

Example:

              1
               \
     1          2
    / \          \
   2   5    =>    3
  / \   \          \
 3   4   6          4
                     \
                      5
                       \
                        6

Basic Idea:

使用分治法。先假设题目所给定的函数可以把一个二叉树变为如上图所示，我们可以先分别用该函数把root.left 和 root.right两棵子树变为这种形式，再将其拼接成最终形式。同时，我们注意到当我们完成左右两子树的flatten之后，如果左子树不存在，我们已经可以直接返回，所以要加上这一判断条件。

Java Code:

    public class Solution {
        /**
         * @param root: a TreeNode, the root of the binary tree
         * @return: nothing
         */

    // 使用分治法解决问题 
        public void flatten(TreeNode root) {
            if (root == null) return;
            flatten(root.left); // flatten 左子树
            flatten(root.right); // flatten 右子树
            if (root.left == null) return;
            TreeNode leftLowest = root.left;
            while (leftLowest.right != null) {
                leftLowest = leftLowest.right;
            }
            leftLowest.right = root.right;
            root.right = root.left;
            root.left = null;
        }
    }

update Aug 26,2017 22:05

更新一段Python的code，与前面先处理再遍历一遍左子树找lowest相比，这种方法虽然代码复杂，但是可以直接返回lowest；

    class Solution:
        """
        @param: root: a TreeNode, the root of the binary tree
        @return: 
        """
        def flatten(self, root):
            def helper(root):
                if not root.left and not root.right:
                    return root
                if not root.left:
                    right = helper(root.right)
                    return right
                if not root.right:
                    left = helper(root.left)
                    root.right = root.left
                    root.left = None
                    return left
                else:
                    left = helper(root.left)
                    right = helper(root.right)
                    left.right = root.right
                    root.right = root.left
                    root.left = None
                    return right

            if root is None:
                return root
            helper(root)
            return root