114.md

114. 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
click to show hints.

Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

my thoughts:

1. dfs recursion.

my solution:

**********44ms
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flatten(self, root):
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """
        if not root:
            return
        def findList(root):
            if not root:
                return [None, None]
            if not root.left and not root.right:
                return [root, root]
            if not root.left:
                return [root, findList(root.right)[1]]
            if not root.right:
                left = findList(root.left)
                root.right = root.left
                root.left = None
                return [root, left[1]]
            
            left = findList(root.left)
            right = findList(root.right)
            root.left = None
            root.right = left[0]
            left[1].right = right[0]
            return [root, right[1]]
            
        findList(root)

my comments:

from other ppl's solution:

1. better recursion:
private TreeNode prev = null;

public void flatten(TreeNode root) {
    if (root == null)
        return;
    flatten(root.right);
    flatten(root.left);
    root.right = prev;
    root.left = null;
    prev = root;
}