Largest Binary Search Subtree in Value - Amazon Top Interview Questions

Problem Statement :

Given a binary tree root, return the largest sum of a subtree that is also a binary search tree.


0 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1


root = [1, [5, [6, null, null], [7, null, null]], [3, [2, null, null], [4, null, null]]]




The binary search tree rooted at 3 has the largest sum with 3 + 2 + 4 = 9

Solution :


                        Solution in C++ :

pair<pair<bool, int>, pair<int, int>> helper(Tree* root, int& res) {
    if (!root) {
        return {{1, 0}, {INT_MAX, INT_MIN}};
    pair<pair<bool, int>, pair<int, int>> leftans = helper(root->left, res);
    pair<pair<bool, int>, pair<int, int>> rightans = helper(root->right, res);

    pair<pair<bool, int>, pair<int, int>> ans;
    ans.first.first = leftans.first.first && rightans.first.first &&
                      root->val > leftans.second.second && root->val < rightans.second.first;
    ans.first.second = leftans.first.second + rightans.first.second + root->val;
    ans.second.first = min({root->val, leftans.second.first, rightans.second.first});
    ans.second.second = max({root->val, leftans.second.second, rightans.second.second});
    if (ans.first.first) {
        res = max(res, ans.first.second);
    return ans;
int solve(Tree* root) {
    int res = 0;
    helper(root, res);
    return res;

                        Solution in Java :

import java.util.*;

 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
class Solution {
    private int maxSubtreeSum = 0;
    static class Node {
        int sum;
        int min;
        int max;
        boolean isBST;
        public Node(int sum, int min, int max, boolean isBST) {
            this.sum = sum;
            this.min = min;
            this.max = max;
            this.isBST = isBST;
    public int solve(Tree root) {
        return maxSubtreeSum;

    private Node dfs(Tree node) {
        if (node.left == null && node.right == null) {
            maxSubtreeSum = Math.max(maxSubtreeSum, node.val);
            return new Node(node.val, node.val, node.val, true);

        int nodeVal = node.val;
        boolean isBST = true;
        int sum, minval, maxval;
        sum = nodeVal;
        minval = nodeVal;
        maxval = nodeVal;

        if (node.left != null) {
            Node lnode = dfs(node.left);
            isBST = (isBST && lnode.isBST && (nodeVal > lnode.max));
            sum += lnode.sum;
            minval = Math.min(minval, lnode.min);
            // maxval = Math.max(maxval, lnode.max);

        if (node.right != null) {
            Node rnode = dfs(node.right);
            isBST = (isBST && rnode.isBST && nodeVal < rnode.min);
            sum += rnode.sum;
            maxval = Math.max(maxval, rnode.max);
            // minval = Math.min(minval, rnode.min);

        if (isBST) {
            maxSubtreeSum = Math.max(maxSubtreeSum, sum);
            return new Node(sum, minval, maxval, true);
        return new Node(sum, minval, maxval, false);

                        Solution in Python : 
class Solution:
    def solve(self, root):
        ans = 0

        def go(root):
            nonlocal ans
            if not root:
                return (0, 1e9, -1e9, 1)

            ls, lmin, lmax, isl = go(root.left)
            rs, rmin, rmax, isr = go(root.right)

            if not isl or not isr or (root.val <= lmax) or (root.val >= rmin):
                return (0, 0, 0, 0)
            ans = max(ans, root.val + ls + rs)
            return (root.val + ls + rs, min(root.val, lmin), max(root.val, rmax), 1)

        return ans

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