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  • [LeetCode] 98. Validate Binary Search Tree

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    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:

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    5
    Input:
    2
    / \
    1 3
    Output: true

    Example 2:

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

    解题报告

    这题主要考查的是二叉搜索树的性质,也就是对于每一个节点来说,它左子树的所有元素都小于它,而右子树的所有元素都大于它。也就是说二叉搜索树在本题中不允许出现两个节点值相同的情况。

    方法一:前序遍历检查是否满足顺序

    二叉搜索树的一个重要特征是,它的前序遍历是符合大小顺序的,因为前序遍历用的是左中右的顺序,这应该就很好理解啦。

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    class Solution:
    def isValidBST(self, root):
    """
    :type root: TreeNode
    :rtype: bool
    """
    self.traverse = []
    self.dfs(root)
    return all(a < b for a, b in zip(self.traverse, self.traverse[1:]))

    def dfs(self, node):
    if not node:
    return
    self.dfs(node.left)
    self.traverse.append(node.val)
    self.dfs(node.right)

    方法二:另一种利用性质的递归

    另一种方法来自于 Leet Code 的讨论区。这种方法的中心思想利用的是节点大小的传递性:对于每个节点来说,它的左子节点的最大值是这个节点的值,而最小值维持和这个节点的最小值相同,右子节点同理。

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

    def check_node(node, l, r):
    if not node:
    return True
    return l < node.val and node.val < r and check_node(node.left, l, node.val) and check_node(node.right, node.val, r)

    class Solution:
    def isValidBST(self, root):
    """
    :type root: TreeNode
    :rtype: bool
    """
    l, r = float("-inf"), float("inf")
    return check_node(root, l, r)