Binary TreesCoding problems

Binary Trees: coding problems

Write a program that reads the input and prints the answer. Run checks it against every test, including hidden ones.

0 of 7 solved

A warehouse stores boxes in a binary tree. Each box holds some number of items, and the tree is written as a line of values in level order: the root first, then each level from left to right. The value -1 marks a missing child. Add up the items in every box.
Read the whole input with cin. Use recursion: an empty tree adds 0, and any other node adds its own value to the sums of its two subtrees.
Constraints: at most 1000 values, each from 0 to 999. The line -1 alone is an empty tree.
Input. One line of level-order values, separated by spaces. Missing children are -1.
Output. One integer: the sum of all values in the tree.
Example 1 input
```
1 2 3 4 5 -1 6
```
Output
```
21
```
Example 2 input
```
-1
```
Output
```
0
```
LanguageC++
#include <iostream> #include <queue> #include <vector> using namespace std; struct TreeNode { int val; TreeNode* left; TreeNode* right; TreeNode(int v) : val(v), left(nullptr), right(nullptr) {} }; TreeNode* build(const vector<int>& a) { if (a.empty() || a[0] == -1) return nullptr; TreeNode* root = new TreeNode(a[0]); queue<TreeNode*> q; q.push(root); size_t i = 1; while (!q.empty() && i < a.size()) { TreeNode* n = q.front(); q.pop(); if (i < a.size() && a[i] != -1) { n->left = new TreeNode(a[i]); q.push(n->left); } i++; if (i < a.size() && a[i] != -1) { n->right = new TreeNode(a[i]); q.push(n->right); } i++; } return root; } int main() { vector<int> values; int x; while (cin >> x) values.push_back(x); TreeNode* root = build(values); // your code here return 0; }