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  • Groupon面经:Find paths in a binary tree summing to a target value

    You are given a binary tree (not necessarily BST) in which each node contains a value. Design an algorithm to print all paths which sum up to that value. Note that it can be any path in the tree - it does not have to start at the root.

    http://stackoverflow.com/questions/4591763/find-paths-in-a-binary-search-tree-summing-to-a-target-value

    Traverse through the tree from the root and do a post-order gathering of all path sums. Use a hashtable at each node to store the possible paths rooted at a node and going down-only. Key is the path sum, value is the actual path. We can construct all paths going through a node from itself and its childrens' paths.

    Here is psuedo-code that implements the above, but stores only the sums and not the actual paths. For the paths themselves, you need to store the end node in the hashtable (we know where it starts, and there's only one path between two nodes in a tree).

    function findsum(tree, target)
      # Traverse the children
      if tree->left
        findsum(tree.left, target)
      if tree->right
        findsum(tree.right, target)
    
      # Single node forms a valid path
      tree.sums = {tree.value}
    
      # Add this node to sums of children
      if tree.left
        for left_sum in tree.left.sums
          tree.sums.add(left_sum + tree.value)
      if tree.right
        for right_sum in tree.right.sums
          tree.sums.add(right_sum + tree.value)
    
      # Have we formed the sum?
      if target in tree.sums
        we have a path
    
      # Can we form the sum going through this node and both children?
      if tree.left and tree.right
        for left_sum in tree.left.sums
          if target - left_sum in tree.right.sums
            we have a path
    
      # We no longer need children sums, free their memory
      if tree.left
        delete tree.left.sums
      if tree.right
        delete tree.right.sums
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5181286.html
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