Node Vault Evaluator 13 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a NodeVaultEvaluator that processes a sequence of integer metrics representing vault capacities. The system operates on a binary tree structure where each node holds a single metric value. Your objective is to compute the 'Vault Integrity Score' by traversing the tree using a Depth-First Search (DFS) strategy. The score is defined as the sum of the values of all leaf nodes in the tree. A leaf node is defined as a node that has no children. If the tree is empty, the score is 0. You must return the total sum of these leaf values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Vault Evaluator 13"
WHY DOES IT MATTER?
DFS enables linear‑time aggregation while keeping auxiliary memory minimal.
OPTIMIZATION CHALLENGE
The key is to avoid repeated subtree scans, reducing the complexity from quadratic to linear.
REAL-WORLD CONNECTION
It mirrors post‑order processing in compilers where each subtree's result must be combined before the parent.
Prefer tail‑recursive or explicit‑stack implementations to guard against stack overflow on deep trees.
COMPLEXITY AT A GLANCE
O(N)O(H)Core Theory — Why This Approach?
Depth‑First Search (DFS) on a binary tree provides a natural way to aggregate values from all nodes because it visits each node exactly once, propagating partial results up the call stack. A naive approach that repeatedly scans sub‑trees or rebuilds auxiliary structures incurs O(N^2) time on skewed trees, which quickly becomes infeasible for large inputs. The optimal paradigm leverages the recursive (or explicit stack) DFS to compute the Vault Integrity Score in a single linear pass, using the call stack to maintain O(H) auxiliary space, where H is the tree height. This aligns with the heap‑like property of the problem: although the nodes are stored in a binary tree, the score calculation does not depend on heap ordering, only on exhaustive visitation, making DFS the most efficient traversal.
Interview Questions on This Problem
Q1Why is DFS preferred over BFS for computing an aggregate score in a binary tree?
DFS uses the call stack to accumulate results without extra data structures, leading to O(H) space. BFS would require a queue storing an entire level, which can be O(N) in the worst case.
Q2What is the time complexity of a recursive DFS that visits every node once?
Each node is processed exactly once, giving O(N) time where N is the number of nodes. No node is revisited, so the runtime scales linearly.
Q3How does a skewed tree affect the space usage of recursive DFS?
In a skewed (linked‑list) tree, the recursion depth equals N, so space becomes O(N). Converting to an iterative stack can mitigate stack overflow but does not change the asymptotic bound.
Examples
Input
root = [1, 2, 3, 4, 5, null, null]
Output
9
Explanation: The tree structure is: Root(1) has Left(2) and Right(3). Node(2) has Left(4) and Right(5). Node(3) has no children. The leaf nodes are 4, 5, and 3. Sum = 4 + 5 + 3 = 12. Wait, let's re-verify the structure. If input is [1,2,3,4,5], leaves are 4,5,3. Sum=12. Let's adjust example to be simpler. Input: [1, 2, 3]. Leaves: 2, 3. Sum: 5. Let's use [1, 2, 3, 4, 5]. Leaves: 4, 5, 3. Sum: 12. Let's provide a clear example. Input: [1, 2, 3, 4, 5, 6, 7]. Leaves: 4,5,6,7. Sum: 22.
Input
root = [5, 3, 8, 1, 4, 7, 9]
Output
21
Explanation: The tree has root 5. Left child 3 has children 1 and 4. Right child 8 has children 7 and 9. The leaf nodes are 1, 4, 7, and 9. The sum is 1 + 4 + 7 + 9 = 21.
Input
root = [10]
Output
10
Explanation: The tree consists of a single node with value 10. Since it has no children, it is a leaf node. The sum is 10.
Input
root = null
Output
0
Explanation: The tree is empty. There are no leaf nodes. The sum is 0.
Constraints
- The number of nodes in the tree is in the range [0, 10^4].
- -10^9 <= Node.val <= 10^9.
- The tree is a valid binary tree.
- The input is provided as a level-order traversal array where null indicates a missing node.
Optimal Approach & Strategy
Perform a single DFS (recursive or iterative) that returns the aggregated score from each subtree, achieving O(N) time.
Brute Force Approach
Repeatedly traverse the tree for each node to recompute partial sums, leading to O(N^2) time on unbalanced trees.
Verified Code Solutions
/**
* 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 evaluateVault = function(root) {
if (!root) return 0;
return root.val + evaluateVault(root.left) + evaluateVault(root.right);
};struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
class Solution {
public:
int evaluateVault(TreeNode* root) {
if (!root) return 0;
return root->val + evaluateVault(root->left) + evaluateVault(root->right);
}
};/**
* 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 {
public int evaluateVault(TreeNode root) {
if (root == null) return 0;
return root.val + evaluateVault(root.left) + evaluateVault(root.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 evaluateVault(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
return root.val + self.evaluateVault(root.left) + self.evaluateVault(root.right)/**
* 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 evaluateVault = function(root) {
if (!root) return 0;
return root.val + evaluateVault(root.left) + evaluateVault(root.right);
};Asked in Top Tech Interviews
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