Path Sum

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Problem statement

You are given the root node of a binary tree consisting of ‘N’ nodes and an integer value ‘TARGET’. Your task is to find the number of leaf nodes in the given binary tree such that the sum of all nodes from the root to leaf is equal to ‘TARGET’.

A leaf node is defined as a node having no child nodes.

You are given a root node ‘ROOT’.Your task is to return the number of leaf nodes satisfying the given condition.

Example: 
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

Example

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation : 
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. 

The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).
Note : 
The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of the input contains an integer, 'T,’ denoting the number of test cases.

The first line of each test case contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example for further clarification.
The second line of each test case contains an integer corresponding to ‘TARGET’.
Output Format: 
For each test case, print the preorder traversal of the given tree.

Print the output of each test case in a separate line.
Note: 
You do not need to print anything. It has already been taken care of. Just implement the given function.
Constraints: 
1 <= T <= 10
1 <= N <= 10^6
1 <= nodeVal <=10^9

Time limit: 1 sec
Sample Input 1:
2
3 1 9 -1 -1 6 10 -1 -1 -1 -1
4
2 1 4 -1 -1 3 -1 -1 -1
11
Sample Output 1:
1
0
Explanation of sample input 1:
For the first test case,

Example

Here, only one path from the root to leaf node [3,1] sum is equal to  ‘TARGET’.Hence, the answer is 1. 

For the second test case,

Example

There is no path from the root to leaf having the sum equal to ‘TARGET’.Hence,  the answer is 0.
Sample Input 2:
2
3 -1 4 -1 10 -1 -1 
17
4 -1 5 5 -1 2 -1 3 -1 -1 -9 -1 -1 
10
Sample Output 2:
1
1
Hint

Try to track the required sum.

Approaches (2)
Using Recursion

This approach will create a recursive function REC(‘CUR’,’REQ_SUM’) that will return the number of good nodes in the subtree with root as ‘CUR’ node and ‘REQ_SUM’ is the required sum. We will first check whether ‘CUR’ node is leaf or not and check the required sum then recursively call this function for its left and right subtree to find the number of required leaf nodes in the left and right subtree, respectively.


 

Algorithm:

  • Defining the REC(‘CUR’,’ REQ_SUM’)
  • If the ‘’CUR’ is an empty node, return 0.
  • If CUR’s left is empty and CUR’s right is empty:
    • CUR is a leaf node.
    • If REQ_SUM is equal to CUR’s value:
      • Return 1.
    • Else:
      • Return 0.
  • Set  ‘ANS’ as 0.
  • Now, update ‘REQ_SUM’ as ‘REQ_SUM’ - CUR’s value.
  • Set ‘ANS’ as ‘ANS’ + REC(left node of ‘CUR’, REQ_SUM).
  • Set ‘ANS’ as ‘ANS’ + REC(right node of ‘CUR’, REQ_SUM).
  • Return ‘ANS’.


 

  • Set ‘ANS’ as REC(‘ROOT’, ‘TARGET’).
  • Return ‘ANS’.
Time Complexity

O(N), where N is the number of nodes in the tree.

 

In this approach, we traverse each node once and check whether the node is leaf with the required sum or not. Hence, the overall time complexity is O(N).

Space Complexity

O(N), where N is the number of nodes in the tree.

 

We use O(N) space complexity for recursive stack during recursion calls in the worst case. The space complexity O(N) is also used for storing the preorder traversal. Hence, the overall space complexity is O(N).

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Path Sum
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