Given a Binary tree and a sum S, print all the paths, starting from root, that sums upto the given sum.
Note that this problem is different from root to leaf paths. Here path doesn’t need to end on a leaf node.
Input : Input : sum = 8, Root of tree 1 / \ 20 3 / \ 4 15 / \ / \ 6 7 8 9 Output : Path: 1 3 4 Input : sum = 38, Root of tree 10 / \ 28 13 / \ 14 15 / \ / \ 21 22 23 24 Output : Path found: 10 28 Path found: 10 13 15
For this problem, preorder traversal is best suited as we have to add up a key value as we land on that node.
We start from the root and start traversing by preorder traversal, adding key value to the sum_so_far and checking whether it is equal to the required sum.
Also, as tree node doesn’t have a pointer pointing to its parent, we have to explicitly save from where we have moved. We use a vector path to store the path for this.
Every node in this path contributes to the sum_so_far.
Path found: 10 28 Path found: 10 13 15
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