LeetCode -- Validate Binary Search Tree

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;
}
}