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Print vertical sum of a binary tree

mahmoud2506
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2022-04-30
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Print vertical sum of a binary tree

 Given a binary tree, find the vertical sum for the binary tree along each vertical line.

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mahmoud2506
2022-04-30
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First we need to understand what vertical sum is. Let go through an example to understand what vertical sum is.

Print vertical sum of a binary tree

For the above tree, let's check what the vertical sum is for the tree.

Print vertical sum of a binary tree 2

Just consider, we have partitioned the tree nodes into column & rows where rows are the level no starting from 0. Then the above tree can be converted into the above table easily.

  1. Start from root. Root column is 0.
  2. If next node is at left of current node
    Then column of next node=column of current node-1
    Else
    Column of next node=column of current node+1

Using the above steps the tree can be easily partitioned to the table. It is to observe that there may be several entry at a specific (row, column) position. Like here, at Level3, column no 1 has two entry 4 & 11.

Rest is about doing sums for each column and printing it.

Thus the vertical sum output should be 2, 12, 8, 20, 9 (from col -2 to col 2 direction).

Though the visual description seems to be very easy to solve this problem, in programing view it’s not that easy.

Algorithm to find vertical sum

The basic concept is to do pre-order traversal & while traversing we will keep track for each column (hashing)& will find cumulative sum.

Thus the column is used as key & we need a map to process our algorithm.

    1. Initialize map<int, int>hash. //(ordered map)
    2. Start from root. Column for root is 0.
      Call vertical_sum(root, 0, hash);
    3. Recursive function
Function vertical_sum( tree node, column, reference of map hash)
a) If (node== NULL)
        Return;
//hash[column]+=node->data
b) Add node value cumulatively for the column; 
// vertical_sum(node->left, column-1, reference of map hash)
c) Recursive do for the left subtree; 
We are passing col-1 since it’s on the left
// vertical_sum(node->right, column+1, reference of map hash)
d) Recursive do for the right subtree; 
We are passing col+1 since it’s on the right

End Functionvertical_sum;
  1. Print the hash map to output the vertical sums for corresponding column.

C++ implementation to print vertical sum of a binary tree

#include <bits/stdc++.h>
using namespace std;

// tree node is defined
class tree{
	public:
		int data;
		tree *left;
		tree *right;
};

//finding vertical sum
void findVerticalSum(tree* root,int col,map<int,int> &hash){ 
	if(root==NULL) //base case
		return;
	//finding sum for respecting column, hashing the column
	hash[col]+=root->data; 
	//recursively process left sub-tree
	findVerticalSum(root->left,col-1,hash); 
	//recursively process right sub-tree
	findVerticalSum(root->right,col+1,hash); 
}

void vertical_sum(tree* root){
	//ordered hash map
	map<int,int> hash; 
	//column no for root is 0
	findVerticalSum(root,0,hash); 
	cout<<"column"<<"\t"<<"sum\n";
	//printing the values from hash map
	for(auto it=hash.begin();it!=hash.end();it++){ 
		//it->first= column no(key) , 
		//it->second=vertical sum of respective column(value)
		cout<<it->first<<"\t"<<it->second<<endl;  
	}
}

// creating new node
tree* newnode(int data)  
{ 
	tree* node = (tree*)malloc(sizeof(tree)); 
	node->data = data; 
	node->left = NULL; 
	node->right = NULL; 
	return(node); 
} 

int main() 
{ 
	//**same tree is builted as shown in example**
	int c,K;
	cout<<"Tree is built like the example aforesaid"<<endl;
	//building the tree like as in the example
	tree *root=newnode(2); 
	root->left= newnode(7); 
	root->right= newnode(5); 
	root->right->right=newnode(9);
	root->right->right->left=newnode(4);
	root->left->left=newnode(2); 
	root->left->right=newnode(6);
	root->left->right->left=newnode(5);
	root->left->right->right=newnode(11);

	cout<<"finding vertical sums......"<<endl; 
	vertical_sum(root);

	return 0; 
}

Output

Tree is built like the example aforesaid
finding vertical sums......
column  sum
-2      2
-1      12
0       8
1       20
2       9

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