[LeetCode] 114. Flatten Binary Tree to Linked List_Medium tag: DFS

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / \
  2   5
 / \   \
3   4   6

The flattened tree should look like:

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

这个题思路就是DFS, 先左后右, 记住如果是用stack如果需要先得到左, 那么要先append右, 另外需要注意的就是用recursive方式的时候需要用一个temp来去存root.right, 否则用preorder的时候root.right就是丢了!

1. Constraints

1) can be empty

2. Ideas

DFS    T: O(n)     S:O(1)

用pre来存之前的node, 然后DFS

3. Code

3.1) iterable

 class Solution:
     def flatten(self, root):
         pre, stack = None, [root]
         while stack:
             node = stack.pop()
             if node:
                 if pre:
                     pre.left = None  # dont forget to set left as None
                     pre.right = node
                 pre = node
                 stack.append(node.right) # stack 先进后出
                 stack.append(node.left)

3.2) recursive

 class Solution:
     def __init__(self):
         self.pre = None
     def flatten(self, root):
         if not root: return
         if self.pre:
             self.pre.left = None
             self.pre.right = root
         self.pre = root
         temp = root.right  # otherwise we will lose the root.right
         self.flatten(root.left)
         self.flatten(root.right)

4. Test cases

1) None

2)

    1
   / \
  2   5
 / \   \
3   4   6