leetcode-114 二叉树展开为链表

给定一个二叉树，原地将它展开为一个单链表。

 

例如，给定二叉树

    1
   / \
  2   5
 / \   \
3   4   6

将其展开为：

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

思路： 对子树进行递归展开，当左右子树都展开完了以后，将左子树接到右子树，再将原来的右子树接上

# 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: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        
        if root is None:
            return 
        
        # 左右子树展开
        self.flatten(root.left)
        self.flatten(root.right)
        

        left = root.left
        right = root.right

        # 将左子树接到根右侧节点
        root.right = left
        root.left = None
        
        # 将原来的右子树接到现在右子树的末尾
        p = root
        while p.right:
            p = p.right
        p.right = right