solve validate-bst

Author: samcho0608

validate-binary-search-tree/samcho0608.java

@@ -0,0 +1,80 @@
+// link: https://leetcode.com/problems/validate-binary-search-tree/description/
+// difficulty: Medium
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution1 {
+    // Problem:
+    // * return: check if tree is a valid BST
+    // * left subtree contains keys strictly less than node's key
+    // * right subtree contains keys strictly greater than node's key
+    // * recursive structure
+    // Solution:
+    // * Time Complexity: O(N)
+    // * Space Complexity(in terms of call stack):
+    //   * O(N) if skewed
+    //   * O(log N) if not skewed
+    public boolean isValidBST(TreeNode root) {
+        // checklist:
+        // * is subtree a valid bst
+        // * left subtree : is max value of subtree less than node
+        // * right subtree : is min value of subtree greater than node
+        return isValid(root, null, null);
+    }
+
+    private boolean isValid(TreeNode node, Integer min, Integer max) {
+        if(node == null) return true;
+
+        int val = node.val;
+
+        if(min != null && val <= min) return false;
+        if(max != null && val >= max) return false;
+
+        if(!isValid(node.left, min, val)) return false;
+        if(!isValid(node.right, val, max)) return false;
+
+        return true;
+    }
+}
+
+// in-order traversal approach
+// * reads from left-most to right (strictly increasing val order expected)
+class Solution2 {
+    private Integer prev;
+
+    // Solution:
+    // * Time Complexity: O(N)
+    // * Space Complexity(in terms of call stack):
+    //   * O(N) if skewed
+    //   * O(log N) if not skewed
+    public boolean isValidBST(TreeNode root) {
+        prev = null;
+        return inorder(root);
+    }
+
+    private boolean inorder(TreeNode node) {
+        // reached end without failure
+        if(node == null) return true;
+
+        // inorder so travel left first
+        if(!inorder(node.left)) return false;
+
+        // if node value is not greater than prev, it isn't a BST in in-order
+        if(prev != null && node.val <= prev) return false;
+        prev = node.val;
+
+        return inorder(node.right);
+    }
+}