Adaptive Target Index — Problem Statement & Solution Guide

TreesMediumDepth-First Search

TimeO(n)

SpaceO(h)

Problem Description

You are given a binary tree where each node contains an integer value. The goal is to compute the 'Adaptive Target Index' for the tree. This index is defined as the sum of all node values in the tree, plus the maximum value found among all nodes. If the tree is empty (i.e., the root is null), the Adaptive Target Index is defined as 0.

The input is the root of a binary tree. Each node has a left child, a right child, and a value. The tree is not necessarily balanced, and node values can be positive, negative, or zero. You must traverse the entire tree to accumulate the total sum and identify the global maximum value.

Return the computed Adaptive Target Index as a single integer. The traversal should be efficient, visiting each node exactly once.

DSA Pattern Breakdown

"Adaptive Target Index"

medium

WHY DOES IT MATTER?

This problem exemplifies the "single‑pass aggregation" pattern, where multiple statistics are derived in one traversal. Mastering this pattern reduces runtime constants, simplifies code, and avoids redundant passes—an essential skill for performance‑critical systems.

OPTIMIZATION CHALLENGE

The key insight is recognizing that sum and maximum are both associative and can be updated independently at each node, allowing them to be merged into a single DFS without extra data structures.

REAL-WORLD CONNECTION

Think of a distributed log processing service that must compute both total traffic volume and peak load from a stream of events. Instead of scanning the log twice, the service aggregates both metrics on the fly, saving I/O and latency—mirroring the tree traversal optimization.

When coding under interview pressure, write a helper that returns a pair (sum, max) from each subtree; this makes the recursion clean, avoids global state, and clearly communicates the combined aggregation to the interviewer.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(h)

Core Theory — Why This Approach?

The Adaptive Target Index (ATI) is a simple aggregate that combines two fundamental tree traversals: computing the total sum of node values and identifying the maximum node value. A naïve solution might perform two separate passes—one to sum all nodes and another to locate the maximum—resulting in O(2n) time, which is still linear but doubles the work and requires extra bookkeeping. More importantly, such an approach often leads to duplicated recursion stacks or iterative loops, increasing the risk of bugs and memory overhead, especially for deep or unbalanced trees.

The optimal paradigm leverages a single depth‑first traversal (pre‑order, in‑order, or post‑order) that aggregates both metrics simultaneously. During the visit of each node, we update a running total and compare the node's value against a current maximum. This yields a true O(n) time solution with O(h) auxiliary space, where h is the tree height, because the recursion stack (or an explicit stack for iterative DFS) is the only extra memory used. By collapsing the two passes into one, we minimize cache misses and keep the algorithm cache‑friendly, which is crucial for large inputs that may contain millions of nodes.

Interview Questions on This Problem

Q1How would you compute the Adaptive Target Index for a binary tree in a single traversal?

Perform a depth‑first search (recursive or iterative). Maintain two variables: totalSum and maxVal. At each node, add its value to totalSum and update maxVal = max(maxVal, node.val). After the traversal, return totalSum + maxVal (or 0 if the tree is empty).

Q2What is the time and space complexity of your solution, and how does it change for a completely balanced versus a skewed tree?

The algorithm visits each node exactly once, giving O(n) time. Space complexity is O(h) due to the recursion stack, which is O(log n) for a balanced tree and O(n) for a completely skewed (linked‑list‑like) tree.

Q3If node values can be negative, does the algorithm need any modification?

No structural change is needed; the same logic works because maxVal is initialized to the smallest possible integer (e.g., Integer.MIN_VALUE) so that even all‑negative trees correctly identify the largest (least negative) value, and the sum correctly accumulates negatives.

Examples

Example 1

Input

root = [5, 3, 8, null, 7, 2, 9]

Output

Explanation: The tree has nodes with values: 5, 3, 8, 7, 2, 9. The sum of all values is 5 + 3 + 8 + 7 + 2 + 9 = 34. The maximum value is 9. The Adaptive Target Index is sum + max = 34 + 9 = 43. Wait, let me recalculate. Sum = 5+3+8+7+2+9 = 34. Max = 9. Result = 34 + 9 = 43. Correction: The example output in the draft was 34, but the logic says sum + max. Let's fix the example to be consistent. Let's use a simpler tree. Root=1, Left=2, Right=3. Sum=6, Max=3, Result=9. Let's create 3 unique examples. Example 1: Tree with root 10, left 5, right 15. Sum = 10+5+15=30. Max=15. Result=45. Example 2: Tree with root -2, left -5, right -1. Sum = -2-5-1=-8. Max=-1. Result=-9. Example 3: Empty tree. Result=0. Example 4: Tree with root 1, left 2, right 3, left-left 4, left-right 5. Sum=1+2+3+4+5=15. Max=5. Result=20. Let's refine the examples to be clear and verified. Example 1: Input: root = [10, 5, 15]. Output: 45. Explanation: Sum = 10+5+15=30. Max=15. 30+15=45. Example 2: Input: root = [-2, -5, -1]. Output: -9. Explanation: Sum = -2-5-1=-8. Max=-1. -8+(-1)=-9. Example 3: Input: root = null. Output: 0. Explanation: Empty tree, return 0. Example 4: Input: root = [1, 2, 3, 4, 5]. Output: 20. Explanation: Sum=1+2+3+4+5=15. Max=5. 15+5=20.

Example 2

Input

root = [-2, -5, -1]

Output

-9

Explanation: The tree contains nodes with values -2, -5, and -1. The sum of all values is -2 + (-5) + (-1) = -8. The maximum value among the nodes is -1. The Adaptive Target Index is calculated as sum + max = -8 + (-1) = -9.

Example 3

Input

root = null

Output

Explanation: The tree is empty. According to the problem definition, if the tree is empty, the Adaptive Target Index is 0.

Example 4

Input

root = [1, 2, 3, 4, 5]

Output

Explanation: The tree has nodes with values 1, 2, 3, 4, and 5. The sum of all values is 1 + 2 + 3 + 4 + 5 = 15. The maximum value is 5. The Adaptive Target Index is 15 + 5 = 20.

Constraints

The number of nodes in the tree is between 0 and 10^5.
-10^9 <= Node.val <= 10^9
The tree is a valid binary tree (each node has at most two children).
The depth of the tree is at most 10^5.

Optimal Approach & Strategy

Combine the two traversals into a single depth‑first search that updates both the running sum and the current maximum at each node, then return their sum.

Brute Force Approach

Run one full traversal to compute the sum of all node values, then run a second independent traversal to find the maximum node value, finally add the two results.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var adaptiveTargetIndex = function(root) {
    if (!root) return 0;
    let sum = 0;
    let maxVal = -Infinity;
    const dfs = (node) => {
        if (!node) return;
        sum += node.val;
        maxVal = Math.max(maxVal, node.val);
        dfs(node.left);
        dfs(node.right);
    };
    dfs(root);
    return sum + maxVal;
};

class Solution {
public:
    int adaptiveTargetIndex(TreeNode* root) {
        if (!root) return 0;
        int sum = 0;
        int maxVal = INT_MIN;
        std::function<void(TreeNode*)> dfs = [&](TreeNode* node) {
            if (!node) return;
            sum += node->val;
            maxVal = std::max(maxVal, node->val);
            dfs(node->left);
            dfs(node->right);
        };
        dfs(root);
        return sum + maxVal;
    }
};

/**
 * 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 Solution {
    private int sum = 0;
    private int maxVal = Integer.MIN_VALUE;
    
    public int adaptiveTargetIndex(TreeNode root) {
        if (root == null) return 0;
        dfs(root);
        return sum + maxVal;
    }
    
    private void dfs(TreeNode node) {
        if (node == null) return;
        sum += node.val;
        maxVal = Math.max(maxVal, node.val);
        dfs(node.left);
        dfs(node.right);
    }
}

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def adaptiveTargetIndex(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        
        def dfs(node):
            nonlocal total_sum, max_val
            if not node:
                return
            total_sum += node.val
            max_val = max(max_val, node.val)
            dfs(node.left)
            dfs(node.right)
            
        total_sum = 0
        max_val = float('-inf')
        dfs(root)
        return total_sum + max_val

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var adaptiveTargetIndex = function(root) {
    if (!root) return 0;
    let sum = 0;
    let maxVal = -Infinity;
    const dfs = (node) => {
        if (!node) return;
        sum += node.val;
        maxVal = Math.max(maxVal, node.val);
        dfs(node.left);
        dfs(node.right);
    };
    dfs(root);
    return sum + maxVal;
};

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Adaptive Target Index Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Adaptive Target Index — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor