Validate Binary Search Tree:一棵树是否是二叉搜索树的检验

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 keysless thanthe node's key.

The right subtree of a node contains only nodes with keysgreater thanthe node's key.

Both the left and right subtrees must also be binary search trees.

Example 1:

    2
   / \
  1   3

Example 2:

    1
   / \
  2   3

思路：二叉搜索树是当前节点比左子树所有节点大且比右子树所有节点小！！！不是比左节点和右节点！！！这点容易写错。

有递归深度优先遍历或者非递归都可以。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null) return true;
        Stack stack = new Stack<>();
        TreeNode pre = null;
        while(root!=null||!stack.isEmpty()){
            while(root!=null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if(pre!=null&&pre.val >= root.val) return false;
            pre = root;
            root = root.right;            
        }
        return true;
    }
}

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