Given a binary tree. The task is to check if the binary tree is sorted level-wise or not. A binary tree is level sorted if max( i-1th level) is less than min( ith level ).
Input : 1 / \ / \ 2 3 / \ / \ / \ / \ 4 5 6 7 Output : Sorted Input: 1 / 4 / \ 6 5 \ 2 Output: Not sorted
Simple Solution: A simple solution is to compare minimum and maximum value of each adjacent level i and i+1. Traverse to ith and i+1th level, compare the minimum value of i+1th level with maximum value of ith level and return the result.
Time complexity: O(n2).
Efficient Solution: An efficient solution is to do level order traversal and keep track of the minimum and maximum values of current level. Use a variable prevMax to store the maximum value of the previous level. Then compare the minimum value of current level with the maximum value of the previous level, pevMax. If minimum value is greater than the prevMax, then the given tree is sorted level-wise up to current level. For next level, prevMax is the equal to maximum value of current level. So update the prevMax with maximum value of current level. Repeat this until all levels of given tree are not traversed.
Below is the implementation of above approach:
Time Complexity: O(n)
Auxiliary Space: O(n)
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