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Find the left view of a binary tree.

by nikoo28
4 comments 3 minutes read

Question: Given the root pointer to a binary tree, find the left view of the tree.

Input: Sample Tree (Pointer to node 1 is given).
Output: 1, 2, 4

At a single glance of the problem, we can understand that we need to perform a level order traversal on the tree. This is similar to finding the number of levels in a tree. At the start of each level, we just print the first element.
This can be done with the algorithm below:-

void leftViewOfBinaryTree(struct binaryTreeNode * root)
	// Level order traversal
	struct binaryTreeNode * temp = NULL;
	struct queue * Q = NULL;
	// Maintain a level count
	int level = 0;

	if(root == NULL)

	Q = enQueue(Q, root);
	//print the root
	printf("%d ",root->data);
	// a flag to check if we need to print
	int needToPrint = 0;
	// Now the first level will end over here,
	// So append a NULL node
	Q = enQueue(Q, NULL);
		temp = Q -> front -> data;
		Q = deQueue(Q);
			printf("%d ", temp->data);
			// toggle the flag
			needToPrint = 0;
		// If we encounter a NULL, that means an end of a level
		// And we need to increment the level count
		if(temp == NULL)
			// Put the marker for next level also
				Q = enQueue(Q, NULL);
			// We are starting a new level, so we will toggle the flag.
			needToPrint = 1;
			// We continue with the level order traversal
			if(temp -> left)
				Q = enQueue(Q, temp -> left);
			if(temp -> right)
				Q = enQueue(Q, temp -> right);
	// Delete the queue

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

Ideone link for the running code:- http://ideone.com/PaJPER

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RaHan May 24, 2016 - 20:54

Your example is incorrect. A level traversal prints 1 2 3 4 5

nikoo28 May 27, 2016 - 10:40

Hi Rahan,

The example is correct. I am printing the left view of the tree. NOT the level order traversal.

Amit March 14, 2015 - 18:58

Why to use an extra storage space – a queue?

void viewLeft(struct treenode *root, int level, int *maxlevel)
	if(root == NULL)
		return ;
	if(*maxlevel info);
		*maxlevel = level;
	viewLeft(root->lchild, level, maxlevel);
	viewLeft(root->rchild, level, maxlevel);

called as viewLeft(root, -1, -1);
TimeComplexity : O(n)

nikoo28 March 16, 2015 - 00:53

Hi Amit,

Thanks for the new approach. The approach discussed in the above post is an iterative one, while you are following a recursive approach. A recursive call back will also maintain a stack of n calls in the worst case. :)

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