计算二叉树有多少个节点

def count(root):
    if not root:
        return 0
   return 1 + count(root.left) + count(root.right)   

计算二叉树的深度

def count_depth(root):
    if not root:
        return 0
    return max(count_depth(root.left), count_depth(root.right)) + 1

leetcode 226 翻转二叉树

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

leetcode 114 二叉树展开为链表

class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return None
        self.flatten(root.left)
        self.flatten(root.right)
        left = root.left
        right = root.right
        root.left = None
        root.right = left
        p = root
        while p.right:
            p = p.right
            
        p.right = right
        return root

【python刷题】二叉树-相关题目

原文：https://www.cnblogs.com/xiximayou/p/14287533.html