Level Order Traversal

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Problem statement

You have been given a Binary Tree of integers. You are supposed to return the level order traversal of the given tree.

For example: 
For the given binary tree

Example

The level order traversal will be {1,2,3,4,5,6,7}.
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.
The first line of each test case contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

Example

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :

Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).
Note : 
The above format was just to provide clarity on how the input is formed for a given tree. 
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Output Format: 
Print all the nodes in level order traversal separated by a single space.
For each test case, print the output in a new line.
Note : 
You do not need to print anything; it has already been taken care of.
Constraints: 
1 <= T <= 100
0 <= N <= 1000
0 <= data <= 10^6 and data != -1
Where ‘T’ is the number of test cases, and ‘N’ is the total number of nodes in the binary tree, and “data” is the value of the binary tree node.

Time Limit: 1sec
Sample Input 1:
3
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
1 2 3 -1 -1 -1 -1
1 3 -1 2 -1 -1 -1
Sample Output 1:
1 2 3 4 5 6 7
1 2 3
1 3 2
Explanation of Sample Input 1:
 For the first test case, {1} is at level 1 and {2,3} are at level 2 and {4,5,6} are at level 3 and {7} is at level 4. So combinations of all levels are {1, 2, 3 ,4, 5, 6, 7}.

For the second test case, {1} is at level 1 and {2,3} are at level 2. So combinations of level are {1, 2, 3}.

For the third test case, {1} is at level 1 and {3} is at level 2 and {2} is at level 3. So combinations of all levels are {1,3,2}.
Sample Input 2:
2
2 7 5 2 6 -1 9 -1 -1 5 11 4 -1 -1 -1 -1 -1 -1 -1
1 2 3 4 -1 5 6 -1 -1 -1 -1 -1 -1
Sample Output 2:
2 7 5 2 6 9 5 11 4
1 2 3 4 5 6
Explanation of Sample Input 2:
For the first test case, {2} is at level 1 and {7,5} are at level 2 and {2,6,9} are at level 3 and {5,11,4} are at level 4. So combinations of all levels are {2, 7, 5, 2, 6, 9, 5, 11, 4}.

For the second test case, {1} is at level 1 and {2,3} are at level 2 and {4,5,6} are at level 3. So combinations of all levels are {1, 2, 3, 4, 5, 6}.
Hint

Which data structure can be used to traverse level by level in the given tree?

Approaches (1)
Breadth First Search

In the level order traversal, we will be using queue data structure which has the property FIRST IN FIRST OUT that’s why which nodes come first in current level the children of that node will also come first for the next level. So, we visit all the nodes one by one of the current level and push into the queue so that when we will be complete with the current level, then we can start exploring nodes of the next level from the queue. We will keep doing until our queue does not become empty and store all the nodes into “output”. Steps are as follows:

  1. Define a queue let’s say as “level”.
  2. Push the root of the given tree into the ”level”.
  3. We will keep doing the following operations until “level” does not become empty:
    • Get the size of the queue, i.e. the total number of nodes at the current level.
    • Visit all the nodes one by one which is at the current level and store values corresponding to that node into “output”. Push their left and right child into the queue if they exist.
  4. Return the “output”.
Time Complexity

O(N), where ‘N’ is the number of nodes in the given binary tree.


We are using the level order traversal, in which there will be ‘N’ push and ‘N’ pop operations, where push and pop operation will take constant time for the queue data structure. So overall time complexity will be O(N).

Space Complexity

O(N), where ‘N’ is the number of nodes in the given binary tree.


We are storing all the nodes for returning the answer; thus, the answer will be the size of ‘N’, and also, we are using the level order traversal. So in the worst case, when the given binary tree will be the balanced binary tree, then the total number of nodes at the bottom level of the tree will be ((N/2)+1). So at the bottom level size of the queue will be the order of O(N). So overall space complexity will be O(N).

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