本文主要是介绍2024.（3.30和4.1）力扣刷题记录-二叉树学习记录2，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
一、学习视频
如何灵活运用递归？【基础算法精讲 10】_哔哩哔哩_bilibili
二、跟练代码
1. 100. 相同的树
递归
# 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 = right
class Solution:def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:# 递归if p is None and q is None:return Trueelif p is None or q is None:return Falseleft = self.isSameTree(p.left,q.left)right = self.isSameTree(p.right,q.right)return left and right and p.val == q.val学习一下灵神的写法，来自视频。
# 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 = right
class Solution:def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:# 递归# if p is None and q is None:# return True# elif p is None or q is None:# return Falseif p is None or q is None:return p is qleft = self.isSameTree(p.left,q.left)right = self.isSameTree(p.right,q.right)return left and right and p.val == q.val2.101. 对称二叉树
递归
# 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 = right
class Solution:def isSymmetric(self, root: Optional[TreeNode]) -> bool:# 递归# 含有根节点# 判断根节点的左右两子树def f(left: Optional[TreeNode], right: Optional[TreeNode]) -> bool:if left is None or right is None:return left is rightreturn f(left.left,right.right) and f(left.right,right.left) and left.val == right.val# #根节点为空时是满足的，此处含根节点可以省去# if root is None: # return Truereturn f(root.left,root.right)3.110. 平衡二叉树
递归。不会，来自视频代码。
# 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 = right
class Solution:def isBalanced(self, root: Optional[TreeNode]) -> bool:# 递归# 计算高度，非负数# 用-1表示不是平衡二叉树def f(node) -> int:if node is None:return 0# left_h = f(node.left)# right_h = f(node.right)# if left_h == -1 or right_h == -1 or abs(left_h - right_h) > 1:# return -1# return max(left_h, right_h) + 1# 少递归一些left_h = f(node.left)if left_h == -1:return -1right_h = f(node.right)if right_h == -1 or abs(left_h - right_h) > 1:return -1return max(left_h, right_h) + 1return f(root) != -14.199. 二叉树的右视图
递归
# 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 = right
class Solution:def rightSideView(self, root: Optional[TreeNode]) -> List[int]:# 递归# 先右后左遍历，当高度大于minHeight时，才会露出来ans = []minHeight = 0def f(node,height):nonlocal minHeightif node is None:return if height > minHeight:ans.append(node.val)minHeight += 1right = f(node.right,height + 1)left = f(node.left,height + 1)return f(root,1)return ans三、课后作业
1.226. 翻转二叉树
递归
# 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 = right
class Solution:def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:# 递归if root is None:return Noneleft = self.invertTree(root.left) #左子树right = self.invertTree(root.right) #右子树root.left = right #左右互换root.right = leftreturn root2.1026. 节点与其祖先之间的最大差值
不会，全都来自灵神题解（. - 力扣（LeetCode）），学习学习。
（1）递
# 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 = right
class Solution:def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:# 递v = -1def f(node,mx,mn):nonlocal vif node is None:return #不一定要有返回值mx = max(mx,node.val)mn = min(mn,node.val)v = max(v, node.val - mn, mx - node.val) #更新f(node.left,mx,mn)f(node.right,mx,mn)f(root, root.val, root.val) #根节点存在return v（2）递的优化
# 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 = right
class Solution:def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:# 递的优化v = -1def f(node,mx,mn):if node is None:nonlocal vv = max(v, mx - mn) #到终点更新returnmx = max(mx,node.val)mn = min(mn,node.val)f(node.left,mx,mn)f(node.right,mx,mn)f(root, root.val, root.val) #根节点存在return v（3）归
# 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 = right
class Solution:def maxAncestorDiff(self, root: Optional[TreeNode]) -> int:# 归v = -1def f(node):if node is None:return inf, -inf # min maxmn = mx = node.vallmin, lmax = f(node.left)rmin, rmax = f(node.right)mn = min(mn, lmin, rmin)mx = max(mx, lmax, rmax)nonlocal v# 每一次都更新vv = max(v, node.val - mn, mx - node.val)return mn, mxf(root)return v递可以不用每次都更新v值，因为从上到下能确保最大小值就是该条路径的，而没有受其他路径影响；而归每次都需更新v值，一个节点有左右两边两条路径到达。
3.1080. 根到叶路径上的不足节点
（1）递归。参考官方题解（. - 力扣（LeetCode））。
# 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 = right
class Solution:def sufficientSubset(self, root: Optional[TreeNode], limit: int) -> Optional[TreeNode]:# 递归# 递总值 归是否删除（False删）def f(node, s):if node is None:return Falseif node.left is None and node.right is None:return s + node.val >= limitleft = f(node.left, s + node.val)right = f(node.right, s + node.val)if not left:node.left = Noneif not right:node.right = Nonereturn left or rightrootBool = f(root, 0)return root if rootBool else None注意删除节点不能本节点直接删除，要在上一节点删除（删除联系）。
（2）递归（调用自身写法）。参考灵神题解（. - 力扣（LeetCode））。
# 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 = right
class Solution:def sufficientSubset(self, root: Optional[TreeNode], limit: int) -> Optional[TreeNode]:# 递归（调用自身写法）# 通过更改limit实现求和的作用if root is None:return Noneif root.left is root.right: #为叶子节点return root if root.val >= limit else Noneroot.left = self.sufficientSubset(root.left, limit - root.val)root.right = self.sufficientSubset(root.right, limit - root.val)return root if root.left or root.right else None可以通过 root.left is root.right 来判断是不是叶子节点。
4.1372. 二叉树中的最长交错路径
（1）递归+非局部变量
# 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 = right
class Solution:def longestZigZag(self, root: Optional[TreeNode]) -> int:# 递归+非局部变量# 递标签，归和值ans = 0def f(node: Optional[TreeNode], flag: str) -> int:nonlocal ansif node is None:return 0left = f(node.left, 'left')right = f(node.right, 'right')if flag == 'left': #该节点是父节点的左节点，下一步该右节点ans = max(ans, left) #更新左节点和值return right+1 #继续归右节点和值else:ans = max(ans, right)return left+1# 更新最终返回值ans = max(f(root.left,"left"),f(root.right,"right"),ans)return ans（2）递归+非局部变量2。来自题解（. - 力扣（LeetCode）），很妙很妙。
# 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 = right
class Solution:def longestZigZag(self, root: Optional[TreeNode]) -> int:# 递归+非局部变量2ans = 0def f(node, l, r): #递左右最大值nonlocal ansans = max(ans, l, r)if node.left:f(node.left, r+1, 0)if node.right:f(node.right, 0, l+1)f(root, 0, 0)return ans完
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