Problem of the day
You have been given a Binary Tree of integers. You are supposed to return the reverse of the level order traversal.
For example:For the given binary tree
The reverse level order traversal will be {7,6,5,4,3,2,1}.
The first line contains elements 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 :
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 reverse level order traversal separated by a single space.
Note :
You do not need to print anything; it has already been taken care of. Just implement the given function.
0 <= N <= 5 * 10^5
0 <= data <= 10^5 and data != -1
Where ‘N’ is the total number of nodes in the binary tree, and 'data' is the value of the binary tree node.
Time Limit: 1 sec
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
7 6 5 4 3 2 1
Start with the bottom of the tree, i.e. 7, and traverse in the tree in the reverse direction of level order, i.e., bottom to top and right to left. So, the reverse level order traversal will be {7,6,5,4,3,2,1}.
2 7 5 2 6 -1 9 -1 -1 5 11 4 -1 -1 -1 -1 -1 -1 -1
4 11 5 9 6 2 5 7 2
Traverse the tree in the reverse direction of level order, i.e., bottom to top and right to left. So reverse level order traversal will be {4,11,5,9,6,2,5,7,2}.
Can you think about exploring the property of the level order traversal?
Our intuition is first to store all the nodes into the vector, let’s say ‘OUTPUT’ while traversing in level by level in the tree.
In level order traversal, we will be going from left to right and top to bottom. So we have stored all the nodes in the left to right and top to bottom manner into the ‘OUTPUT’. But our desired traversal is “from right to left and bottom to top”. So if we reverse our stored nodes, i.e. ‘OUTPUT’ then left to right will change into the right to left and top to bottom will change into the bottom to top.
The steps are as follows:
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, the overall time complexity will be O(N).
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, the overall space complexity will be O(N).