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'.
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 :

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.
1 <= T <= 2*10^2
0 <= N <= 3*10^3
0 <= NODE.DATA <= 10^5
Time Limit : 1 sec
1
5 6 2 4 3 -1 -1 9 7 -1 -1 -1 -1 -1 -1
False

The leaves (2, 7) and (2, 9) are two levels apart, thus the answer is False.
2
2 3 4 -1 -1 -1 -1
8 -1 -1
True
True
Try to save the level of one leaf and compare the level of all the other leaves with this level.
The main idea behind this approach is to keep track of the level of the current node (using the variable say, âcurLevelâ) and the minimum and the maximum levels of the leaf nodes encountered (in the variables say, âminLeafLevelâ and âmaxLeafLevelâ respectively). We traverse the entire tree recursively, starting from the root node. Initially, âcurLevelâ, âminLeafLevelâ and âmaxLeafLevelâ, all are set to 0. As we move down the tree, we increment the value of âcurLevelâ. When we encounter the first leaf, we set the values of both âminLeafLevelâ and âmaxLeafLevelâ equal to that of âcurLevelâ.
Now, as we traverse the tree, whenever a leaf is encountered we do the following :
O(N) , where âNâ is the number of nodes in the tree.
In the worst case, we traverse the entire tree once. Hence, the overall time complexity will be O(N).
O(N), where âNâ is the number of nodes in the tree.
The recursive stack will take O(H) space where âHâ is the height of the tree. In the worst case, âHâ will equal to âNâ, when given tree is skewed tree. Hence, the overall space complexity will be O(N).