Question Link:

https://leetcode.com/problems/flatten-binary-tree-to-linked-list/

Dynamic Programming

Traverse the binary tree pre-orderly, and keep track the last visit node. For each visit node, link the last visit node with the current node (set left to NULL and right to the current node).

Python implementation

The python code is as follows. The code is based on the solution of [LeetCode 94] Binary Tree Inorder Traversal.

[-] Python code accepted by LeetCode OJ 
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def flatten(self, root):
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """

        if root is None:
            return root

        # Stack for pre-order traverse
        S = [root]

        # Previous visited node
        prev_node = TreeNode(None)

        # Preorder traverse
        while S:
            N = S.pop()

            # Visit
            prev_node.left = None
            prev_node.right = N

            # Now N is the previously visited node
            prev_node = N

            # Push right which will be visited later
            if N.right:
                S.append(N.right)
            # Push left which will be visited earlier
            if N.left:
                S.append(N.left)

        # Set the last node
        prev_node.left = None
        prev_node.right = None

Extend Reading

There is another post Convert a Binary Tree to a Circular Doubly Link List, which is a more complicated problem.