LeetCode -- Validate Binary Search Tree

Posted 2020-06-17

Question:

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.

 

Analysis:

给出一棵二叉树，判断它是否是二叉搜索树（BST）。

假设BST是这样定义的：它要么是一棵空树，要么是具有以下性质的树：

1）左子树的所有结点的关键码小于根节点的关键码；

2）右子树的所有结点的关键码大于根节点的关键码；

3）左子树和右子树也是二叉搜索树。

 

思路：对于一棵BST来说，它的中序遍历的值正好是从大到小排列的，因此可以根据中序遍历来判断二叉树是否是BST。

 

Answer：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        List<Integer> l = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        do {
            while(p != null) {
                stack.push(p);
                p = p.left;
            }
            if(!stack.isEmpty()) {
                p = stack.pop();
                if(l.size() != 0) {
                    if(p.val <= l.get(l.size()-1))
                        return false;
                }
                l.add(p.val);
                p = p.right;
            }
        } while(p != null || !stack.isEmpty());
        return true;
    }
}