题解 | #判断一棵二叉树是否为搜索二叉树和完全二叉树# 一次递归遍历同时判断

非常直观的代码，直接看代码就能懂

class Solution:
    def judgeIt(self , root: TreeNode) -> List[bool]:
        
        from dataclasses import dataclass
        
        @dataclass
        class Info:
            mx: int  # 整棵树的最大值
            mi: int  # 整棵树的最小值
            height: int    # 树的高度
            is_bst: bool   # 是否搜索二叉树
            is_full: bool  # 是否满二叉树
            is_cbt: bool   # 是否完全二叉树
        
        def dfs(x):
            if not x: return Info(float('-inf'), float('inf'), 0, True, True, True)
            
            l, r = dfs(x.left), dfs(x.right)
            # 使用左右子树的信息得到当前节点的信息
            mx = max(x.val, r.mx)
            mi = min(x.val, l.mi)
            height = max(l.height, r.height) + 1
            is_bst = l.is_bst and r.is_bst and l.mx < x.val < r.mi
            is_full = l.is_full and r.is_full and l.height == r.height
            is_cbt = is_full or \
                l.is_full and r.is_full and l.height - 1 == r.height or \
                l.is_full and r.is_cbt and l.height == r.height or \
                l.is_cbt and r.is_full and l.height - 1 == r.height
            
            return Info(mx, mi, height, is_bst, is_full, is_cbt)
            
        info = dfs(root)
        return info.is_bst, info.is_cbt