Skip to content
Related Articles

Related Articles

Find maximum vertical sum in binary tree
  • Difficulty Level : Medium
  • Last Updated : 26 Oct, 2020

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: 



C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find maximum vertical
// sum in binary tree.
#include <bits/stdc++.h>
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<pair<Node*, int> > 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
                  \
                   
    */
 
    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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find maximum
# vertical sum in binary tree.
from sys import maxsize
from collections import deque
 
INT_MIN = -maxsize
 
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
 
# Function to find maximum vertical sum
# in binary tree.
def maxVerticalSum(root: Node) -> int:
     
    if (root is None):
        return 0
 
    # To store sum of each vertical level.
    verSum = dict()
 
    # To store maximum vertical level sum.
    maxSum = INT_MIN
 
    # To store vertical level of current node.
    currLev = 0
 
    # Queue to perform level order traversal.
    # Each element of queue is a pair of node
    # and its vertical level.
    q = deque()
    q.append([root, 0])
 
    while (q):
 
        # Extract node at front of queue
        # and its vertical level.
        root = q[0][0]
        currLev = q[0][1]
        q.popleft()
 
        # Update vertical level sum of
        # vertical level to which
        # current node belongs to.
        if currLev not in verSum:
            verSum[currLev] = 0
             
        verSum[currLev] += root.data
 
        if (root.left):
            q.append([root.left, currLev - 1])
 
        if (root.right):
            q.append([root.right, currLev + 1])
 
    # Find maximum vertical level sum.
    for it in verSum:
        maxSum = max([maxSum, verSum[it]])
 
    return maxSum
 
# Driver code
if __name__ == "__main__":
     
    '''
                3
              /  \
             4    6
           /  \  /  \
         -1   -2 5   10
                  \
                   
    '''
 
    root = Node(3)
    root.left = Node(4)
    root.right = Node(6)
    root.left.left = Node(-1)
    root.left.right = Node(-2)
    root.right.left = Node(5)
    root.right.right = Node(10)
    root.right.left.right = Node(8)
 
    print(maxVerticalSum(root))
 
# This code is contributed by sanjeev2552

chevron_right


Output: 

14




 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :