验证二叉搜索树

Recursive! Recursive! And Recursive!

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.

I can't solve that😅

主要是边界的问题一直没解决然后超时了。。。

Discussion

class Solution {
    public boolean helper(TreeNode node, Integer lower, Integer upper) {
        if (node == null)
            return true;

        int val = node.val;
        if (lower != null && val <= lower)
            return false;
        if (upper != null && val >= upper)
            return false;

        if (!helper(node.right, val, upper))
            return false;
        if (!helper(node.left, lower, val))
            return false;
        return true;
    }

    public boolean isValidBST(TreeNode root) {
        return helper(root, null, null);
    }
}

官方答案，很妙，但是凭空我是想不起来的。

//如果不用递归，而用栈实现前序：

class Solution {
    Stack<TreeNode> st = new Stack<>();
    Stack<Long> upperList = new Stack<>(), 
        lowerList = new Stack<>();
        
    public boolean isValidBST(TreeNode root) {
        long lower = Long.MIN_VALUE, upper = Long.MAX_VALUE, val;
        update(root, lower, upper);
        while (!st.empty()) {
            root = st.pop();
            lower = lowerList.pop();
            upper = upperList.pop();
            if (root == null) continue;
            val = (long)root.val;
            if (val <= lower) return false;
            if (val >= upper) return false;
            update(root.right, val, upper);
            update(root.left, lower, val);
        }
        return true;
    }

    void update(TreeNode node, long lower, long upper) {
        st.push(node);
        lowerList.push(lower);
        upperList.push(upper);
    }
}