leetcode: 树 98. Validate Binary Search Tree

题目

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

Example 1:

    2
   / \
  1   3

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

思路

思路1： 用递归，每次判断 root-left 和root-right的大小，然后给左子树和右子树设一个取值范围。用递归判断子树是否为二叉搜索树。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidSub(self,root,base_left,base_right):
        if root.left:
            if root.left.val >= root.val or root.left.val <= base_left:
                return False
            if not self.isValidSub(root.left,base_left,root.val):
                return False
            
        if root.right:
            if root.right.val <= root.val or root.right.val >=base_right:
                return False
            if not self.isValidSub(root.right,root.val,base_right):
                return False
            
        return True
    
    def isValidBST(self, root: TreeNode) -> bool:
        if root == None:
            return True
        
        return self.isValidSub(root,-sys.maxsize,sys.maxsize)

思路二：

用栈，按照 left-root-right的顺序遍历树，如果遍历到的数字要比前一个数字小，则返回 False。如果遍历完都正常，则返回True

    def isValidBST(self, root: TreeNode) -> bool:
        if root == None:
            return True
        
        stack_root = []
        nums = [-sys.maxsize]
        while(1):
            while(1):
                while root.left:
                    if root.left.val >= root.val:
                        return False
                    stack_root.append(root)
                    root = root.left
                if root.val <= nums[-1]:
                    return False
                nums.append(root.val)
                if root.right:
                    if root.right.val <= root.val:
                        return False
                    root = root.right
                else:
                    break
                
            if len(stack_root)>0:
                root = stack_root.pop()
                if root.val <= nums[-1]:
                    return False
                nums.append(root.val)
                while (root.right == None) and len(stack_root)>0:
                    root = stack_root.pop()
                    if root.val <= nums[-1]:
                        return False
                    nums.append(root.val)
                if root.right:
                    root = root.right
                else:
                    break
            else:
                break
        return True