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算法第十七天｜654. 最大二叉树、617.合并二叉树、700.二叉搜索树中的搜索、98.验证二叉搜索树

backend 2025/5/14 7:50:04

654. 最大二叉树

题目

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思路与解法

# 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 constructMaximumBinaryTree(self, nums: List[int]) -> Optional[TreeNode]:if not nums:return Nonemax_val = max(nums)max_idx = nums.index(max_val)root = TreeNode(max_val)root.left = self.constructMaximumBinaryTree(nums[:max_idx])root.right = self.constructMaximumBinaryTree(nums[max_idx+1:])return root

617.合并二叉树

题目

在这里插入图片描述

思路与解法

第一想法： 通过层序遍历来合并
一遍过，哈哈，虽然肯定有可以优化的地方

# 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 mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:if not root1:return root2if not root2:return root1from collections import dequequeue1 = deque()queue2 = deque()queue1.append(root1)queue2.append(root2)while queue1 or queue2:cur1 = queue1.popleft()cur2 = queue2.popleft()cur_new_val = cur1.val + cur2.valcur1.val = cur_new_valif not cur1.left and cur2.left:cur1.left = TreeNode(0)elif cur1.left and not cur2.left:cur2.left = TreeNode(0)if not cur1.right and cur2.right:cur1.right = TreeNode(0)elif cur1.right and not cur2.right:cur2.right = TreeNode(0)if cur1.left:queue1.append(cur1.left)queue2.append(cur2.left)if cur1.right:queue1.append(cur1.right)queue2.append(cur2.right)return root1

700.二叉搜索树中的搜索

题目

在这里插入图片描述

思路与解法

第一想法：

# 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 searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:val_node = self.find_root(root, val)return val_nodedef find_root(self, node, val):if not node:return Noneif node.val == val:return nodeelif node.val > val:return self.find_root(node.left, val)elif node.val < val:return self.find_root(node.right, val)

98.验证二叉搜索树

题目

在这里插入图片描述

思路与解法

第一想法： 判断每个结点作为root结点构成的子树是不是二叉搜索树，如下。但是有坑，光判断每个子数是不够的，因为子树中结点可能不满足子树外的条件

# 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 isValidBST(self, root: Optional[TreeNode]) -> bool:if not root:return Trueif root.left and root.left.val >= root.val:return Falseif root.right and root.right.val <= root.val:return Falsereturn self.isValidBST(root.left) and self.isValidBST(root.right)

如下就不满足条件，
在这里插入图片描述

思考后的想法： 中序遍历后，挨个判断是不是递增的。发现carl也是这么做的，只是他有几种方法。

# 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 isValidBST(self, root: Optional[TreeNode]) -> bool:self.nodes = []self.traversal(root)for i in range(len(self.nodes)-1):if self.nodes[i] >= self.nodes[i+1]:return Falsereturn Truedef traversal(self, root):if not root:return Noneself.traversal(root.left)self.nodes.append(root.val)self.traversal(root.right)