02. Leaf under budget
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The problem can be found at the following link:
I am using a breadth-first search (BFS) approach to traverse the tree.
I maintain a queue to keep track of nodes and their corresponding levels.
As I traverse the tree level by level, I check if a node is a leaf node (both left and right children are null).
If it is a leaf node and the level is greater than the budget (k), I return the count of leaf nodes found so far.
Otherwise, I increment the count and subtract the current level from the budget.
If the node has children, I push them into the queue with their corresponding levels.
In this solution, as we traverse from level 1 down to there leaf nodes, it naturally addresses our greedy approach to maximize the number of leaf nodes within our budget, providing the desired answers automatically.
Time Complexity: The algorithm visits each node in the tree once, so the time complexity is O(N)
, where N
is the number of nodes in the tree.
Auxiliary Space Complexity: The space complexity is O(W)
, where W
is the maximum number of nodes at any level in the tree.
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