Last Updated: 8 Oct, 2020

Leaves One Level Apart

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Problem statement

You are given a binary tree of 'N' nodes. Your task is to print 'True' if the leaves of the tree are at most one level apart else print 'False'.

Input Format :

The first line of input contains an integer 'T' representing the number of test cases. Then the test cases follow.

The only line of each test case 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 on its place.

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

Example Input

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 :

For each test case, print "True" if the leaves of the tree are at most one level apart else print "False".

The output of each test case will be printed in a separate line.

Note :

You don't need to print anything. It has already been take care of, just implement the given function.

Constraints :

1 <= T <= 2*10^2
0 <= N <= 3*10^3
0 <= NODE.DATA <= 10^5

Time Limit : 1 sec

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Approaches

01 Approach

The main idea behind this approach is to keep track of the level of the current node (using the variable say, ‘curLevel’) and the minimum and the maximum levels of the leaf nodes encountered (in the variables say, ‘minLeafLevel’ and ‘maxLeafLevel’ respectively). We traverse the entire tree recursively, starting from the root node. Initially, ‘curLevel’, ‘minLeafLevel’ and ‘maxLeafLevel’, all are set to 0. As we move down the tree, we increment the value of ‘curLevel’. When we encounter the first leaf, we set the values of both ‘minLeafLevel’ and ‘maxLeafLevel’ equal to that of ‘curLevel’.

Now, as we traverse the tree, whenever a leaf is encountered we do the following :

We update the value of ‘minLeafLevel’ to ‘curLevel’ if the value of ‘curLevel’ is less than that of ‘minLeafLevel’.
We update the value of ‘maxLeafLevel’ to ‘curLevel’ if the value of ‘curLevel’ is more than that of ‘maxLeafLevel’.
Now, we check the difference between the following variables :
- ‘minLeafLevel’ and ‘curLevel’
- ‘maxLeafLevel’ and ‘curLevel’
If any of the above two differences is more than one, we have found a pair of leaf nodes that are more than one level apart. In this case, we return “False”.
Else, we move to the next node.
We return “True” if after the entire traversal no pair of leaf nodes is found that are more than one level apart.

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