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Find all root to leaf path sum of a Binary Tree
  • Difficulty Level : Medium
  • Last Updated : 12 Jan, 2021

Given a Binary Tree, the task is to print all the root to leaf path sum of the given Binary Tree.

Examples: 

Input:
                 30
              /      \
            10        50
           /  \      /  \
          3   16   40    60
Output: 43 56 120 140
Explanation:
In the above binary tree
there are 4 leaf nodes. 
Hence, total 4 path sum are
present from root node to the
leaf node. 
(i.e., 30-10-3, 30-10-16,
       30-50-40, 30-50-60) 
Therefore, the path sums from
left to right would be
(43, 56, 120, 140).

Input:
                 11
              /      \
            12        5
              \      /  
              16   40
Output: 39 56
Explanation:
In the above binary tree
there are 2 leaf nodes. 
Hence, total 2 path sum are
present from root node to
the leaf node.
(i.e 11-12-16, 11-5-40) 
Therefore, the path sums from
left to right would be (39 56).

Approach: The idea is to use DFS Traversal to travel from the root to the leaf of the binary tree and calculate the sum of each root to leaf path. Follow the steps below to solve the problem:

  1. Start from the root node of the Binary tree with the initial path sum of 0.
  2. Add the value of the current node to the path sum.
  3. Travel to the left and right child of the current node with the present value of the path sum.
  4. Repeat Step 2 and 3 for all the subsequent nodes of the binary tree.
  5. On reaching the leaf node, add the path sum to the vector of pathSum.
  6. Print all the elements of the vector pathSum as the output.

Below is the implementation of the above approach: 

C++14

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// TreeNode structure
struct TreeNode {
    int val;
    TreeNode *left, *right;
};
 
// Function that returns a new TreeNode
// with given value
struct TreeNode* addNode(int data)
{
    // Allocate memory
    struct TreeNode* node
        = (struct TreeNode*)malloc(
            sizeof(struct TreeNode));
 
    // Assign data to val variable
    node->val = data;
    node->left = NULL;
    node->right = NULL;
    return node;
};
 
// Function that calculates the root to
// leaf path sum of the BT using DFS
void dfs(TreeNode* root, int sum,
         vector<int>& pathSum)
{
    // Return if the node is NULL
    if (root == NULL)
        return;
 
    // Add value of the node to
    // the path sum
    sum += root->val;
 
    // Store the path sum if node is leaf
    if (root->left == NULL
        and root->right == NULL) {
 
        pathSum.push_back(sum);
        return;
    }
 
    // Move to the left child
    dfs(root->left, sum, pathSum);
 
    // Move to the right child
    dfs(root->right, sum, pathSum);
}
 
// Function that finds the root to leaf
// path sum of the given binary tree
void findPathSum(TreeNode* root)
{
    // To store all the path sum
    vector<int> pathSum;
 
    // Calling dfs function
    dfs(root, 0, pathSum);
 
    // Printing all the path sum
    for (int num : pathSum) {
        cout << num << " ";
    }
    cout << endl;
}
 
// Driver Code
int main()
{
    // Given binary tree
    TreeNode* root = addNode(30);
    root->left = addNode(10);
    root->right = addNode(50);
    root->left->left = addNode(3);
    root->left->right = addNode(16);
    root->right->left = addNode(40);
    root->right->right = addNode(60);
 
    /*  The above code constructs this tree
 
                30
             /      \
           10        50
          /  \      /  \
         3   16   40    60
   */
 
    // Function Call
    findPathSum(root);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Structure of
// binary tree node
static class Node
{
    int val;
    Node left, right;
};
 
// Function to create new node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.val = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function that calculates the root to
// leaf path sum of the BT using DFS
static void dfs(Node root, int sum,
                ArrayList<Integer> pathSum)
{
     
    // Return if the node is NULL
    if (root == null)
        return;
 
    // Add value of the node to
    // the path sum
    sum += root.val;
 
    // Store the path sum if node is leaf
    if (root.left == null &&
       root.right == null)
    {
        pathSum.add(sum);
        return;
    }
 
    // Move to the left child
    dfs(root.left, sum, pathSum);
 
    // Move to the right child
    dfs(root.right, sum, pathSum);
}
 
// Function that finds the root to leaf
// path sum of the given binary tree
static void findPathSum(Node root)
{
     
    // To store all the path sum
    ArrayList<Integer> pathSum = new ArrayList<>();
 
    // Calling dfs function
    dfs(root, 0, pathSum);
 
    // Printing all the path sum
    for(int num : pathSum)
    {
        System.out.print(num + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Construct binary tree
    Node root = newNode(30);
    root.left = newNode(10);
    root.right = newNode(50);
    root.left.left = newNode(3);
    root.left.right = newNode(16);
    root.right.left = newNode(40);
    root.right.right = newNode(60);
 
    /*  The above code constructs this tree
 
            30
         /      \
       10        50
      /  \      /  \
     3   16   40    60
*/
 
    // Function call
    findPathSum(root);
}
}
 
// This code is contributed by offbeat

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Python3

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# Python3 program for the
# above approach
from collections import deque
 
# A Tree node
class Node:
   
    def __init__(self, x):
       
        self.data = x
        self.left = None
        self.right = None
         
pathSum = []
 
# Function that calculates
# the root to leaf path
# sum of the BT using DFS
def dfs(root, sum):
   
    # Return if the node
    # is NULL
    if (root == None):
        return
 
    # Add value of the node to
    # the path sum
    sum += root.data
 
    # Store the path sum if
    # node is leaf
    if (root.left == None and
        root.right == None):
        pathSum.append(sum)
        return
 
    # Move to the left child
    dfs(root.left, sum)
 
    # Move to the right child
    dfs(root.right, sum)
 
# Function that finds the
# root to leaf path sum
# of the given binary tree
def findPathSum(root):
   
    # To store all the path sum
    # vector<int> pathSum
 
    # Calling dfs function
    dfs(root, 0)
 
    # Printing all the path sum
    for num in pathSum:
        print(num, end = " ")
         
# Driver Code
if __name__ == '__main__':
   
    # Given binary tree
    root = Node(30)
    root.left = Node(10)
    root.right = Node(50)
    root.left.left = Node(3)
    root.left.right = Node(16)
    root.right.left = Node(40)
    root.right.right = Node(60)
 
   #  /*  The above code constructs this tree
   #
   #              30
   #           /      \
   #         10        50
   #        /  \      /  \
   #       3   16   40    60
   # */
 
    # Function Call
    findPathSum(root)
 
# This code is contributed by Mohit Kumar 29

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
 
// Structure of
// binary tree node
public class Node
{
    public int val;
    public Node left, right;  
    public Node(int item)
    {
        val = item;
        left = right = null;
    }
}
public class BinaryTree
{
    public static Node node;
   
    // Function that calculates the root to
    // leaf path sum of the BT using DFS
    static void dfs(Node root, int sum, List<int> pathSum)
    {
       
        // Return if the node is NULL
        if (root == null)
        {
            return;
        }
       
        // Add value of the node to
        // the path sum
        sum += root.val;
       
        // Store the path sum if node is leaf
        if(root.left == null && root.right == null)
        {
            pathSum.Add(sum);
            return;
        }
       
        // Move to the left child
        dfs(root.left, sum, pathSum);
       
        // Move to the right child
        dfs(root.right, sum, pathSum);
    }
   
    // Function that finds the root to leaf
    // path sum of the given binary tree
    static void findPathSum(Node root)
    {
       
        // To store all the path sum
        List<int> pathSum = new List<int>();
       
        // Calling dfs function
        dfs(root, 0, pathSum);
       
        // Printing all the path sum
        foreach(int num in pathSum)
        {
            Console.Write(num + " ");
        }
    }
   
    // Driver code
    static public void Main ()
    {
       
        // Construct binary tree
        BinaryTree.node = new Node(30);
        BinaryTree.node.left = new Node(10);
        BinaryTree.node.right = new Node(50);
        BinaryTree.node.left.left = new Node(3);
        BinaryTree.node.left.right = new Node(16);
        BinaryTree.node.right.left = new Node(40);
        BinaryTree.node.right.right = new Node(60);
         
      /*  The above code constructs this tree
  
            30
         /      \
       10        50
      /  \      /  \
     3   16   40    60
*/
        // Function call
        findPathSum(node);
    }
}
 
// This code is contributed by avanitrachhadiya2155

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Output: 

43 56 120 140

 

Time Complexity: O(N)
Auxiliary Space: O(N) 

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