/*
This is an implementation that determines
if a given binary tree is a binary search
tree.
In a binary search tree, for ever node v:
- Elements in left subtree rooted at v
are less than element stored at v.
- Elements in right subtree rooted at v
are greater than or equal to one at v.
Let n be the number of nodes in the
binary search tree.
Time complexity: O(n)
Space complexity: O(n)
*/
public class ValidBST {
private BTNode BTRoot;
public ValidBST() {
/*
* Create tree below:
* * 4
* /
* 2 7
* * /
* * 6 8
*/
BTRoot = new BTNode(4, null, null);
BTNode rootLeft = new BTNode(2, null, null);
BTRoot.left = rootLeft;
BTNode rootRight = new BTNode(7, null, null);
BTRoot.right = rootRight;
BTNode rootRightLeft = new BTNode(6, null, null);
BTNode rootRightRight = new BTNode(8, null, null);
rootRight.left = rootRightLeft;
rootRight.right = rootRightRight;
}
public void setRootVal(int val) {
BTRoot.val = val;
}
public static void main(String[] args) {
ValidBST application = new ValidBST();
// The considered Binary Tree is a BST
System.out.println(application.validateBST()); // true
// Change root value such that result Binary Tree
// is no longer a BST.
application.setRootVal(1);
System.out.println(application.validateBST()); // false
}
public boolean validateBST() {
return isValidBST(BTRoot, Integer.MIN_VALUE, Integer.MAX_VALUE);
}
private static boolean isValidBST(BTNode root, int min, int max) {
// Empty trees are valid BSTs
if (root == null) {
return true;
}
// The current value must be between min and max
if (root.val < min || root.val >= max) {
return false;
}
// Current value is max for left subtree
// and min for right subtree.
return isValidBST(root.left, min, root.val) &&
isValidBST(root.right, root.val, max);
}
// Class representing a binary tree node
// with pointers to value, left, and right nodes
private class BTNode {
int val;
BTNode left;
BTNode right;
public BTNode(int val, BTNode left, BTNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
}