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Diagonal Traversal of Binary Tree

In this article, we are going to see how to print diagonal traversal of a binary tree? This problem has been featured in coding round of Amazon.
Submitted by Radib Kar, on February 20, 2019

Problem statement:

Given a binary tree, print the diagonal traversal of the binary tree.

Example:

Consider lines of slope -1 passing between nodes. Given a Binary Tree, print all diagonal elements in a binary tree belonging to same line.

Diagonal Traversal of Binary Tree

In the above example lines of slope -1 are passed between nodes and the diagonal traversal will be: 2 5 9 7 6 11 4 2 5

Solution

Pre-requisite:

Queue q, root of binary tree, Node* temp, Node* temp1

Algorithm:

The algorithm is actually processing the right children and EnQueueing the left children for each parent node. The EnQueued children accts as nodes to be processed for next level.

FUNCTION diagonalPrint(root)   
EnQueue (q, root);
	While (q is not empty){
		temp=DeQueue(q);
		Print temp->data;
		temp1=temp;
		While (temp1 is not NULL)
			IF (temp1->left is not NULL) //if left child exists
			EnQueue (q, temp1->left); //enqueue
			END IF
			temp1=temp1->right; //traverse to right child
			IF (temp1 is not NULL) 
			Print temp1->data;
			END IF
		END While
	END While
END FUNCTION

Example with explanation:

Nodes are represented with their respective values for better understanding.

In main we call diagonalPrint(2)
---------------------------------------------------------

diagonalPrint(2)
q.push(2)
---------------------------------------------------------

1st iteration
temp=2
print 2
temp1=temp//2

////////////////////inner while loop////////////////////
0th iteration
temp1 is not NULL
2->left is not NULL
q.push(2->left) //7
temp1=2->right//5
Print 5
1st iteration
temp1 is not NULL//5
temp1->left is NULL
nothing to push
temp1=temp1->right //9
Print 9
2nd iteration
temp1 is not NULL//9
temp1->left is not NULL
q.push(9->left)//4
temp1=temp1->right //NULL
nothing to print
3rd iteration
temp1 NULL
exit from inner loop
////////////////////inner while loop ends////////////////////

Queue status:
7 4
---------------------------------------------------------

2nd iteration
temp=q.pop() =7
print 7
temp1=temp//7

////////////////////inner while loop////////////////////
0th iteration
temp1 is not NULL
7->left is not NULL
q.push(7->left) //2
temp1=7->right//6
Print 6
1st iteration
temp1 is not NULL//6
temp1->left is not NULL
q.push(6->left) //5
temp1=temp1->right //11
Print 11
2nd iteration
temp1 is not NULL//11
temp1->left is NULL
nothing to push
temp1=temp1->right //NULL
nothing to print
3rd iteration
temp1 NULL
exit from inner loop
////////////////////inner while loop ends////////////////////

Queue status:
4 2 5
---------------------------------------------------------

3rd iteration
temp=q.pop()=4
print 4
temp1=temp//4

////////////////////inner while loop////////////////////
0th iteration
temp1 is not NULL
4->left is NULL
Nothing to push
temp1=4->right//NULL

1st iteration
temp1 is NULL
exit from inner loop
////////////////////inner while loop ends////////////////////

Queue status:
2 5
---------------------------------------------------------

4th iteration
temp=q.pop()=2
print 2
temp1=temp//2

////////////////////inner while loop////////////////////
0th iteration
temp1 is not NULL
2->left is NULL
Nothing to push
temp1=2->right//5
Print 5
1st iteration
temp1 is not NULL//5
temp1->left is NULL
nothing to push
temp1=temp1->right //NULL

2nd iteration
temp1 NULL
exit from inner loop
////////////////////inner while loop ends////////////////////

Queue status:
Empty queue
---------------------------------------------------------
Program ends.
Hence the diagonal traversal is:
 2 5 9 7 6 11 4 2 5

C++ implementation:

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

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

void diagonalPrint(Node *root)
{
	Node* temp,*temp1;
	queue<Node*> q;
	q.push(root);
	while(!q.empty()){
		temp=q.front();
		cout<<temp->data<<" ";
		q.pop();
		temp1=temp;
		while(temp1){
			//if left child exists EnQueue
			if(temp1->left) 
				q.push(temp1->left);
			
			//process all right children
			temp1=temp1->right; 
			if(temp1)
				cout<<temp1->data<<" ";
		}
	}
}

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

int main() 
{ 
	//**same tree is builted as shown in example**
	Node *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<<"Diagonal traversal of binary tree is :"<<endl; 
	diagonalPrint(root); 

	return 0; 
}

Output

Diagonal traversal of binary tree is :
2 5 9 7 6 11 4 2 5





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