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LargestBSTSubtree.java
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184 lines (158 loc) · 6.73 KB
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/**
* Given a binary tree,
* find the largest subtree which is a Binary Search Tree (BST),
* where largest means subtree with largest number of nodes in it.
*/
public class LargestBSTSubtree {
public static class LargestBST {
public Node node;
public int maxNode;
public int min;
public int max;
public LargestBST(Node n, int number, int min_value, int max_value) {
node = n;
max = max_value;
min = min_value;
maxNode = number;
}
} // end of class LargestBST
public static void main(String[] args) {
Node root = new Node(10);
Node node5 = new Node(5);
Node node15 = new Node(15);
Node node6 = new Node(6);
Node node7 = new Node(7);
node15.right = node7;
Node node3 = new Node(3);
Node node12 = new Node(12);
Node node11 = new Node(11);
Node node13 = new Node(13);
Node node14 = new Node(14);
root.insert(node5);
root.insert(node6);
root.insert(node15);
root.insert(node3);
root.insert(node12);
root.insert(node11);
root.insert(node13);
root.insert(node14);
int result = largestBSTSubtree(root);
System.out.println("largest bst (may not include all children): " + result);
LargestBST largeBST = largestBSTSubtree1(root);
if (largeBST != null)
System.out.println(largeBST.node.value + " : size " + largeBST.maxNode);
LargestBST largeBST2 = largestBSTSubtree2(root);
if (largeBST2 != null)
System.out.println(largeBST2.node.value + " : size " + largeBST2.maxNode);
}
// Given a binary tree, find the largest Binary Search Tree (BST),
// where largest means BST with largest number of nodes in it.
// The largest BST may or may not include all of its descendants.
public static int largestBSTSubtree(Node node) {
if (node == null)
return 0;
if (node.left == null && node.right == null)
return 1;
int leftNode = largestBSTSubtree(node.left);
int rightNode = largestBSTSubtree(node.right);
if (node.left != null && node.right != null) {
if ((node.left.value < node.value) && (node.right.value > node.value)) {
return leftNode + rightNode + 1;
} else if (node.left.value < node.value) {
return leftNode + 1;
} else if (node.right.value > node.value) {
return rightNode + 1;
} else {
return Math.max(rightNode, leftNode);
}
} else if (node.left != null) {
if (node.left.value < node.value)
return leftNode + 1;
else
return leftNode;
} else {// if (node.right != null){
if (node.value < node.right.value)
return rightNode + 1;
else
return rightNode;
}
}
// Given a binary tree, find the largest Binary Search Tree (BST),
// where largest means BST with largest number of nodes in it.
// The largest BST may or may not include all of its descendants.
public static LargestBST largestBSTSubtree1(Node node) {
if (node == null)
return null;
if (node.left == null && node.right == null) {
return new LargestBST(node, node.size(), node.value, node.value);
}
LargestBST leftNode = largestBSTSubtree1(node.left);
LargestBST rightNode = largestBSTSubtree1(node.right);
if (leftNode != null && rightNode != null) {
if ((node.value > leftNode.max && node.left == leftNode.node)
&& (node.value < rightNode.min && node.right == rightNode.node)) {
LargestBST bst = new LargestBST(node,
leftNode.maxNode + rightNode.maxNode + 1,
leftNode.min,
rightNode.max);
return bst;
} else if (node.value > leftNode.max && node.left == leftNode.node) {
return new LargestBST(node, leftNode.maxNode + 1, leftNode.min, node.value);
} else if (node.value < rightNode.min && node.right == rightNode.node) {
return new LargestBST(node, rightNode.maxNode + 1, node.value, rightNode.max);
} else {
return (leftNode.maxNode > rightNode.maxNode) ? leftNode : rightNode;
}
} else if (leftNode != null) {
if (node.value > leftNode.max && node.left == leftNode.node) {
return new LargestBST(node, leftNode.maxNode + 1, leftNode.min, node.value);
} else {
return leftNode;
}
} else if (rightNode != null) {
if (node.value < rightNode.min && node.right == rightNode.node) {
return new LargestBST(node, rightNode.maxNode + 1, node.value, rightNode.max);
} else {
return rightNode;
}
}
return null;
}
// Given a binary tree, find the largest Binary Search Tree (BST),
// where largest means BST with largest number of nodes in it.
// The largest BST must include all of its descendants.
public static LargestBST largestBSTSubtree2(Node node) {
if (node == null)
return null;
if (node.left == null && node.right == null) {
return new LargestBST(node, node.size(), node.value, node.value);
}
LargestBST leftNode = largestBSTSubtree2(node.left);
LargestBST rightNode = largestBSTSubtree2(node.right);
if (leftNode != null && rightNode != null) {
if ((node.value > leftNode.max && node.left == leftNode.node)
&& (node.value < rightNode.min && node.right == rightNode.node)) {
LargestBST bst = new LargestBST(node,
leftNode.maxNode + rightNode.maxNode + 1,
leftNode.min,
rightNode.max);
return bst;
} else {
return (leftNode.maxNode > rightNode.maxNode) ? leftNode : rightNode;
}
} else if (leftNode != null) {
if (node.value > leftNode.max && node.left == leftNode.node) {
return new LargestBST(node, leftNode.maxNode + 1, leftNode.min, node.value);
} else {
return leftNode;
}
} else if (rightNode != null) {
if (node.value < rightNode.min && node.right == rightNode.node) {
return new LargestBST(node, rightNode.maxNode + 1, node.value, rightNode.max);
} else {
return rightNode;
}
}
return null;
}
}