# Find maximum vertical sum in binary tree

Given a binary tree, find the maximum vertical level sum in binary tree.

Examples:

```Input :
3
/  \
4    6
/  \  /  \
-1   -2 5   10
\
8

Output : 14
Vertical level having nodes 6 and 8 has maximum
vertical sum 14.

Input :
1
/  \
5    8
/  \    \
2   -6    3
\       /
-1     -4
\
9

Output : 4
```

A simple solution is to first find vertical level sum of each level starting from minimum vertical level to maximum vertical level. Finding sum of one vertical level takes O(n) time. In worst case time complexity of this solution is O(n^2).

An efficient solution is to do level order traversal of given binary tree and update vertical level sum of each level while doing the traversal. After finding vertical sum of each level find maximum vertical sum from these values.

Below is the implementation of above approach:

 `// CPP program to find maximum vertical ` `// sum in binary tree. ` `#include ` `using` `namespace` `std; ` ` `  `// A Binary Tree Node ` `struct` `Node { ` `    ``int` `data; ` `    ``struct` `Node *left, *right; ` `}; ` ` `  `// A utility function to create a new ` `// Binary Tree Node ` `struct` `Node* newNode(``int` `item) ` `{ ` `    ``struct` `Node* temp = (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node)); ` `    ``temp->data = item; ` `    ``temp->left = temp->right = NULL; ` `    ``return` `temp; ` `} ` ` `  `// Function to find maximum vertical sum ` `// in binary tree. ` `int` `maxVerticalSum(Node* root) ` `{ ` `    ``if` `(root == NULL) { ` `        ``return` `0; ` `    ``} ` ` `  `    ``// To store sum of each vertical level. ` `    ``unordered_map<``int``, ``int``> verSum; ` ` `  `    ``// To store maximum vertical level sum. ` `    ``int` `maxSum = INT_MIN; ` ` `  `    ``// To store vertical level of current node. ` `    ``int` `currLev; ` ` `  `    ``// Queue to perform level order traversal. ` `    ``// Each element of queue is a pair of node ` `    ``// and its vertical level. ` `    ``queue > q; ` `    ``q.push({ root, 0 }); ` ` `  `    ``while` `(!q.empty()) { ` ` `  `        ``// Extract node at front of queue ` `        ``// and its vertical level. ` `        ``root = q.front().first; ` `        ``currLev = q.front().second; ` `        ``q.pop(); ` ` `  `        ``// Update vertical level sum of ` `        ``// vertical level to which ` `        ``// current node belongs to. ` `        ``verSum[currLev] += root->data; ` ` `  `        ``if` `(root->left) ` `            ``q.push({ root->left, currLev - 1 }); ` ` `  `        ``if` `(root->right) ` `            ``q.push({ root->right, currLev + 1 }); ` `    ``} ` ` `  `    ``// Find maximum vertical level sum. ` `    ``for` `(``auto` `it : verSum)  ` `        ``maxSum = max(maxSum, it.second); ` `    `  `    ``return` `maxSum; ` `} ` ` `  `// Driver Program to test above functions ` `int` `main() ` `{ ` `    ``/* ` `                ``3 ` `              ``/  \ ` `             ``4    6 ` `           ``/  \  /  \ ` `         ``-1   -2 5   10 ` `                  ``\ ` `                   ``8   ` `    ``*/` ` `  `    ``struct` `Node* root = newNode(3); ` `    ``root->left = newNode(4); ` `    ``root->right = newNode(6); ` `    ``root->left->left = newNode(-1); ` `    ``root->left->right = newNode(-2); ` `    ``root->right->left = newNode(5); ` `    ``root->right->right = newNode(10); ` `    ``root->right->left->right = newNode(8); ` ` `  `    ``cout << maxVerticalSum(root); ` `    ``return` `0; ` `} `

Output:

```14
```

Time Complexity: O(n)
Auxiliary Space: O(n)

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