sinat_40469883的博客

题目![在这里插入图片描述](https://img-blog.csdnimg.cn/20210710235529582.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3NpbmF0XzQwNDY5ODgz,size_16,color_FFF# Definition for a binary tree node.# class TreeNode:

# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def flatten(self, root: TreeNode) ->...

# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclass Solution: def diameterOfBinaryTree(self, root: Tr...

思路：dfs函数用来判断B中所有节点是否与A中当前节点开始的数相同，然后主函数判断B是不是A的子结构# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def isSubStructure..

# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNod...

思路：dfs + 回溯注意：如果语句写在递归下面就是回溯，当二叉树遍历完左子树回去的时候就会删除节点# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def pathSum(se..

class Solution: def verifyPostorder(self, postorder: List[int]) -> bool: def helper(left, right): if left >= right: return True p = left while postorder[p] < postorder[right]: p += 1 ...

出错的例子：出错原因：递归前两个条件没有return，调用函数无输出接口，所以最后return Noneclass Solution: def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': def CommonAncestor(root, p, q): valr = root.val v.

递归：由上至下递归，需要传递路径这个参数，递归函数无返回class Solution: def binaryTreePaths(self, root: TreeNode) -> List[str]: def allPaths(root, path): if root: path += str(root.val) if not root.left and not root.right:

思路和二叉树的最大深度不同，二叉树的最大深度只需要返回max(左子树最大深度，右子树最大深度)+1就可以二叉树的最小深度需要判断其左子树和右子树是否存在class Solution: def minDepth(self, root: TreeNode) -> int: if not root: return 0 if not root.left and not root.right: return 1

自底向上的递归class Solution: def isBalanced(self, root: TreeNode) -> bool: def height(root): if not root: return 0 height_left = height(root.left) height_right = height(root.right)

目录用递归和广度优先搜索（BFS）实现二叉树的最大深度递归：自底向上（比较容易理解）递归：自顶向下广度优先搜索，思路与二叉树的层序遍历相同用递归和广度优先搜索（BFS）实现二叉树的最大深度递归：自底向上（比较容易理解）class Solution: def maxDepth(self, root: TreeNode) -> int: def depth(root): if not root: .