Complexity appears to be O(n lgn), but actually O(n)

Author: idf

src/BalancedBinaryTree/Solution.java

@@ -14,7 +14,9 @@
  */
 public class Solution {
     /**
-     * O(n lgn)
+     * O(n lgn)?
+     * Actually it is O(n)
+     * \sum_{i=0}^{h-1} 2^i (h-i)
      * @param root
      * @return
      */

src/ValidateBinarySearchTree/Solution.java

@@ -0,0 +1,45 @@
+package ValidateBinarySearchTree;
+
+import commons.datastructures.TreeNode;
+
+/**
+ * User: Danyang
+ * Date: 1/27/2015
+ * Time: 20:52
+ *
+ * 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.
+ */
+public class Solution {
+    /**
+     * Complexity appears to be O(n lgn), but actually O(n)
+     * @param root
+     * @return
+     */
+    public boolean isValidBST(TreeNode root) {
+        if(root==null)
+            return true;
+        TreeNode l = getMax(root.left);
+        TreeNode r = getMin(root.right);
+        if(l!=null && l.val>=root.val || r!=null && r.val<=root.val)
+            return false;
+        return isValidBST(root.left) && isValidBST(root.right);
+    }
+
+    TreeNode getMax(TreeNode cur) {
+        while(cur!=null && cur.right!=null)
+            cur = cur.right;
+        return cur;
+    }
+
+    TreeNode getMin(TreeNode cur) {
+        while(cur!=null && cur.left!=null)
+            cur = cur.left;
+        return cur;
+    }
+}