98. Validate Binary Search Tree

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

Binary tree [2,1,3], return true.

Example 2:

    1
   / 
  2   3

Binary tree [1,2,3], return false.

本题采用的方法是迭代的先序遍历检验法，因为起点变成了从最左节点开始，所以可以当作没有左节点，只要比较其父节点和父节点的右子树节点大小即可，stack用来存储其父节点。代码如下：

/**
 * 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) {
        TreeNode node = root;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode pre = null;
        while(node!=null||!stack.isEmpty()){
            while(node!=null){
                stack.push(node);
                node = node.left;
            }
            node = stack.pop();
            if(pre!=null&&pre.val>=node.val) return false;
            pre = node;
            node = node.right;
        }
        return true;
    }
}