LeetCode98 validate-binary-search-tree（验证二叉搜索树）

题目

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

A valid 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: 
Input: root = [2,1,3]
Output: true

 Example 2: 
Input: root = [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.

 Constraints: 
 The number of nodes in the tree is in the range [1, 10⁴]. 
 -2³¹ <= Node.val <= 2³¹ - 1 

方法

递归法

• 时间复杂度：O(n)
• 空间复杂度：O(n)

class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidBST(root,Long.MIN_VALUE,Long.MAX_VALUE);
    }
    private boolean isValidBST(TreeNode root,long lower,long upper) {
        if(root==null){
            return true;
        }
        if(root.val<=lower || root.val>=upper){
            return false;
        }
        return isValidBST(root.left, lower, root.val) && isValidBST(root.right, root.val, upper);
    }
}

中序遍历法

根据二叉搜索树的特性，其中序遍历是单调递增的，因此只要在中序遍历的过程中判断是否单调递增即可

• 时间复杂度：O(n)
• 空间复杂度：O(n)

class Solution {
    public boolean isValidBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        long max = Long.MIN_VALUE;
        while(!stack.isEmpty()||root!=null){
            while(root!=null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if(root.val<=max){
                return false;
            }
            max = root.val;
            root = root.right;
        }
        return true;
    }
}