Given a binary tree, count leaves in the tree without using recursion. A node is a leaf node if both left and right children of it are NULL.
Leaves count for the above tree is 3.
The idea is to use level order traversal. During traversal, if we find a node whose left and right children are NULL, we increment count.
Time Complexity: O(n)
Here is a recursive solution for the same problem:
The idea is simple we brake the larger tree into smaller sub-trees and solve for them to get the final answer.
Below is the implementation of the above approach.
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