Leetcode-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.

验证一颗二叉树是否是搜索二叉树。（不允许重复键的二叉树）
标签：深度优先搜索。

思路：对于二叉树的深度优先搜索实际上就是先序遍历（preorder traversal）和中序遍历（inorder traversal），两者的入栈和出栈顺序是一样的，区别在于：遍历到节点A时，先序遍历先访问节点A，再遍历A的左子节点；中序遍历则先遍历左子节点，出栈后再访问节点A。
由于对于搜索二叉树来说，有一个特殊的性质，即中序遍历后的结果是升序的（这里需要注意，如果允许重复键的话，搜索二叉树要求左子树的键不大于根节点的键，此时无法直接用中序遍历来验证是否是搜索二叉树，因为无法区分l<= root < r和 l < root <= r）。

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private long prev = (long)Integer.MIN_VALUE - 1;

public boolean isValidBST(TreeNode root) {
return inorder(root);
}

private boolean inorder(TreeNode h) {
if (h == null)
return true;
if (!inorder(h.left))
return false;
if (h.val > prev)
prev = h.val;
else
return false;
return inorder(h.right);
}
}