Print all k-sum paths in a binary tree - GeeksforGeeks

Print all k-sum paths in a binary tree - GeeksforGeeks
A binary tree and a number k are given. Print every path in the tree with sum of the nodes in the path as k.
A path can start from any node and end at any node, i.e. they need not be root node and leaf node; and negative numbers can also be there in the tree.

The basic idea to solve the problem is to do a preorder traversal of the given tree. We also need a container (vector) to keep track of the path that led to that node. At each node we check if there are any path that sums to k, if any we print the path and proceed recursively to print each path.

http://hevi.info/interviews/print-all-k-sum-paths-in-a-binary-tree/

My approach is basic, first think about a Tree with root only, if its value is equal to the given value then it is a solution. Now lets add a left and a right node. In this case as we go deeper on the tree we need to look for;

• starting from root, is the sum equals to the given value
• starting from the left is the sum equals to the given value and same for right

Hard part is collecting the correct path information.

The BST in solution hold lower values on the left and BST may contain negative values.

The complexity of my solution is O(n^2) hope it is not a concern for you

void findSum(Integer originalSum, Integer sum,

             List<List<Integer>> result, List<Integer> currentList,

             Node node) {

 

    if (node == null) return;

 

    Integer nodeValue = node.value;

    currentList.add(nodeValue);

 

    if (Objects.equals(sum, nodeValue)) {

        List<Integer> resultL = new ArrayList(currentList);

        result.add(resultL);

    }

 

    findSum(originalSum, originalSum, result, new ArrayList(), node.right);

 

    findSum(originalSum, originalSum, result, new ArrayList(), node.left);

 

    int remaining = sum - nodeValue;

    findSum(originalSum, remaining, result, new ArrayList(currentList), node.left);

 

    findSum(originalSum, remaining, result, new ArrayList(currentList), node.right); 

}

void printKPathUtil(Node *root, vector<int>& path,

                                           int k)

{

    // empty node

    if (!root)

        return;

 

    // add current node to the path

    path.push_back(root->data);

 

    // check if there's any k sum path

    // in the left sub-tree.

    printKPathUtil(root->left, path, k);

 

    // check if there's any k sum path

    // in the right sub-tree.

    printKPathUtil(root->right, path, k);

 

    // check if there's any k sum path that

    // terminates at this node

    // Traverse the entire path as

    // there can be negative elements too

    int f = 0;

    for (int j=path.size()-1; j>=0; j--)

    {

        f += path[j];

 

        // If path sum is k, print the path

        if (f == k)

            printVector(path, j);

    }

 

    // Remove the current element from the path

    path.pop_back();

}

 

// A wrapper over printKPathUtil()

void printKPath(Node *root, int k)

{

    vector<int> path;

    printKPathUtil(root, path, k);

}

http://algorithms.tutorialhorizon.com/print-paths-in-binary-tree-with-sumx/
Given a binary tree and X, Print all the paths start­ing from root so that sum of all the nodes in path equals to a given num­ber.
public void hasPath(Node root, int sum, String path){
  if(root!=null){
    if(root.data>sum){ // if root is greater than Sum required, return
      return;
    }else{
      path+=" " + root.data; //add root to path
      sum=sum-root.data; // update the required sum
      if(sum==0){ //if sum required =0, means we have found the path
        System.out.println(path);
      }
      hasPath(root.left, sum, path);
      hasPath(root.right, sum, path);
    }
  }
http://www.geeksforgeeks.org/print-paths-root-specified-sum-binary-tree/

oid printPathsUtil(Node* curr_node, int sum,

             int sum_so_far, vector<int> &path)

{

    if (curr_node == NULL)

        return;

 

    // add the current node's value

    sum_so_far += curr_node->key;

 

    // add the value to path

    path.push_back(curr_node->key);

 

    // print the required path

    if (sum_so_far == sum )

    {

        cout << "Path found: ";

        for (int i=0; i<path.size(); i++)

            cout << path[i] << " ";

 

        cout << endl;

    }

 

    // if left child exists

    if (curr_node->left != NULL)

        printPathsUtil(curr_node->left, sum, sum_so_far, path);

 

    // if right child exists

    if (curr_node->right != NULL)

        printPathsUtil(curr_node->right, sum, sum_so_far, path);

 

 

    // Remove last element from path

    // and move back to parent

    path.pop_back();

}

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