Find maximum vertical sum in binary tree in C++


Suppose we have a binary tree. The task is to print maximum of the sum of all nodes in the vertical order traversal. So if the tree is like below −

The vertical order traversal is like −

4
2
1 + 5 + 6 = 12
3 + 8 = 11
7
9

Here the maximum is 12. The approach is simple. We will perform the vertical order traversal, then find the sum and check for maximum.

Example

#include<iostream>
#include<map>
#include<vector>
#include<queue>
using namespace std;
class Node {
   public:
   int key;
   Node *left, *right;
};
Node* getNode(int key){
   Node* node = new Node;
   node->key = key;
   node->left = node->right = NULL;
   return node;
}
int getMaxSum(Node* root) {
   if (!root)
   return -1;
   int n = 0;
   int k_node = -1;
   map<int, vector<int> > current_map;
   int hd = 0;
   queue<pair<Node*, int> > que;
   que.push(make_pair(root, hd));
   while (!que.empty()) {
      pair<Node*, int> temp = que.front();
      que.pop();
      hd = temp.second;
      Node* node = temp.first;
      current_map[hd].push_back(node->key);
      if (node->left != NULL)
      que.push(make_pair(node->left, hd - 1));
      if (node->right != NULL)
      que.push(make_pair(node->right, hd + 1));
   }
   map<int, vector<int> >::iterator it;
   int maximum = INT_MIN;
   for (it = current_map.begin(); it != current_map.end(); it++) {
      int temp_sum = 0;
      for (int i = 0; i < it->second.size(); ++i) {
         temp_sum += it->second[i];
      }
      if(maximum < temp_sum){
         maximum = temp_sum;
      }
   }
   return maximum;
}
int main() {
   Node* root = getNode(1);
   root->left = getNode(2);
   root->right = getNode(3);
   root->left->left = getNode(4);
   root->left->right = getNode(5);
   root->right->left = getNode(6);
   root->right->right = getNode(7);
   root->right->left->right = getNode(8);
   root->right->right->right = getNode(9);
   cout << "Maximum sum of vertical nodes: " << getMaxSum(root);
}

Output

Maximum sum of vertical nodes: 12

Updated on: 18-Dec-2019

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