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1. Use BFS traversal (use queue)
- rparundekar March 02, 20102. at each node currentSum = sum obtained from parent + current value
3. if the currentSum is the required sum and node is leaf node, you have got solution
4. else if currentSum > required sum, do not add child
5. else
5.a Add left child to the queue;
5.b add right child to queue only if (requiredSum-currentSum<=currentSum)
BFS ensures minimum solution. 5.b and 4 steps prune search space