代码随想录算法训练营第17天

654.最大二叉树

# 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]:
        max_val=0
        for i in range(len(nums)):
            max_val=max(max_val,nums[i])
        if not nums:
            return 
        root=TreeNode(max_val)
        index=nums.index(max_val)
        root.left=self.constructMaximumBinaryTree(nums[:index])
        root.right=self.constructMaximumBinaryTree(nums[index+1:])
        return root

617.合并二叉树

# 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 root1 is None:
            return root2
        if root2 is None:
            return root1
        root1.val+=root2.val
        
        root1.left=self.mergeTrees(root1.left,root2.left)
        root1.right=self.mergeTrees(root1.right,root2.right)

        return root1

700.二叉搜索树中的搜索

# 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]:
        #利用二叉搜索树的特性来做这道题
        if not root or root.val==val:
            return root
        if root.val > val:
            return self.searchBST(root.left,val)
        if root.val < val:
            return self.searchBST(root.right,val)
        

98.验证二叉搜索树

为什么要用中序？因为只有中序（左中右）的顺序才能满足大小顺序。
第一种办法是将二叉树转换成数组，然后看数组是否是顺序排列。我们此处使用中序遍历的方法解决这道题

# 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 __init__(self):
        self.max_value=float('-inf')
    def isValidBST(self, root: Optional[TreeNode]) -> bool:
        if root is None:
            return True
        left=self.isValidBST(root.left)
        if root.val > self.max_value:
            self.max_value=root.val
        else:
            return False
        right=self.isValidBST(root.right)
        return left and right  #只有当左子树和右子树都是二叉搜索树返回True