validate binary search tree

Change-Id: I0cbf47e0226bb55bf57e9f8890acbb6ebffc09aa

Author: applewjg

ValidateBinarySearchTree.java

@@ -0,0 +1,52 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 20, 2014
+ Problem:    Validate Binary Search Tree
+ Difficulty: Medium
+ Source:     https://oj.leetcode.com/problems/validate-binary-search-tree/
+ Notes:
+ 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.
+
+ Solution: Recursion. 1. Add lower & upper bound. O(n)
+                      2. Inorder traversal with one additional parameter (value of predecessor). O(n)
+ */
+
+/**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    boolean isValidBSTRe(TreeNode root, long left, long right)
+    {
+        if(root == null) return true;
+        return left < root.val && root.val < right &&
+                isValidBSTRe(root.left,left,root.val) 
+                && isValidBSTRe(root.right, root.val, right);
+    }
+    public boolean isValidBST_1(TreeNode root) {
+        if (root == null) return true;
+        return isValidBSTRe(root, (long)Integer.MIN_VALUE - 1, (long)Integer.MAX_VALUE + 1);
+    }
+    boolean isValidBST(TreeNode root) {
+        long[] val = new long[1];
+        val[0] = (long)Integer.MIN_VALUE - 1;
+        return inorder(root, val);
+    }
+    boolean inorder(TreeNode root, long[] val) {
+        if (root == null) return true;
+        if (inorder(root.left, val) == false) 
+            return false;
+        if (root.val <= val[0]) return false;
+        val[0] = root.val;
+        return inorder(root.right, val);
+    }
+}