Find paths in a binary search tree summing to a target value

Question

Given a binary search tree and a target value, find all the paths (if there exists more than one) which sum up to the target value. It can be any path in the tree. It doesn\'t h

Answer 1

My answer is O(n^2), but I believe it is accurate, and takes a slightly different approach and looks easier to implement.

Assume the value stored at node i is denoted by VALUE[i]. My idea is to store at each node the sum of the values on the path from root to that node. So for each node i, SUM[i] is sum of path from root to node i.

Then for each node pair (i,j), find their common ancestor k. If SUM(i)+SUM(j)-SUM(k) = TARGET_SUM, you have found an answer.

This is O(n^2) since we are looping over all node pairs. Although, I wish I can figure out a better way than just picking all pairs.

We could optimize it a little by discarding subtrees where the value of the node rooted at the subtree is greater than TARGET_SUM. Any further optimizations are welcome :)

Psuedocode:

# Skipping code for storing sum of values from root to each node i in SUM[i]
for i in nodes:
    for j in nodes:
        k = common_ancestor(i,j)
        if ( SUM[i] + SUM[j] - SUM[k] == TARGET_SUM ):
            print_path(i,k,j)

Function common_ancestor is a pretty standard problem for a binary search tree. Psuedocode (from memory, hopefully there are no errors!):

sub common_ancestor (i, j):
  parent_i = parent(i)
  # Go up the parent chain until parent's value is out of the range. 
  # That's a red flag.
  while( VAL[i] <= VAL[parent_i] <= VAL[j] ) : 
    last_parent = parent_i
    parent_i = parent(i)
    if ( parent_i == NULL ): # root node
      break
return last_parent