98. Validate Binary Search Tree

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98. Validate Binary Search Tree

题目描述

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.

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Example 1:
    2
   / \
  1   3
Binary tree [2,1,3], return true.
Example 2:
    1
   / \
  2   3
Binary tree [1,2,3], return false.

题目大意

校验是否是二叉搜索树。

校验规则：

左子树的所有节点值都比根节点小；

右子树的所有节点值都比根节点大；

左子树和右子树也是二叉搜索树

解题思路

DFS，根据题目所述的规则进行判定即可。

代码

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func isValidBST(_ root: TreeNode?) -> Bool {
        
        guard let tree = root else { return true }
        
        var ans = true
        var rootVal = tree.val
        
        func dfs(_ t: TreeNode, min: Int, max: Int) {
            
            if let left = t.left {
                //是否小于根节点，通过max判定左子树是否都小于根节点
                if left.val >= t.val || left.val <= min || left.val >= max {
                    ans = false
                    return
                }
                dfs(left, min: min, max: t.val)
            }
            
            if let right = t.right {
                //是否大于根节点，通过min判定右子树是否都大于根节点
                if right.val <= t.val || right.val <= min || right.val >= max {
                    ans = false
                    return
                }
                dfs(right, min: t.val, max: max)
            }
        }
        
        dfs(tree, min: Int.min, max: Int.max)
        return ans
    }
}
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