Java Program For Sum Root to Leaf Numbers

Sum Root to Leaf Numbers is a popular interview problem which is asked in companies like Google & Microsoft. You can find this question in LeetCode & InterviewBit.

We have been given a binary tree where each node can contain any number between 0-9.
    2
/
4 8
For the above tree, there are two root to leaf paths, 2->4 = 24 & 2->8 = 28. So result should be 24 + 28 =  52. In InterviewBit problem, we need to get the remainder after dividing the result by 1003 (result % 1003) to restrict number overflow. Number overflow can happen if tree has a lot of depth.
Below you can find the working Java solution for this problem. It will do a pre-order traversal & do a backtracking to find all possible solutions & add them to result. Time complexity is O(n) as we need to traverse the full tree once. Space complexity is constant O(1) if we don’t consider recursion stack.

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import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;

//Definition for binary tree
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
left=null;
right=null;
}
}

class Solution {
public int sumNumbers(TreeNode A) {
String currentSum = "";
//Like Integer, BigInteger object is also immutable. Assigning new object refrence
//inside method won't change method argument. So using an array of BigInteger.
BigInteger[] result = new BigInteger[]{BigInteger.valueOf(0)};
calculateSumRootToLeaf(A, currentSum, result);
result[0] = result[0].remainder(BigInteger.valueOf(1003));
return result[0].intValue();
}
private void calculateSumRootToLeaf(TreeNode node, String currentSum, BigInteger[] result){
currentSum = currentSum + "" + node.val;
if(node.left == null && node.right == null){
result[0] = result[0].add(new BigInteger(currentSum));
} else {
if(node.left != null){
calculateSumRootToLeaf(node.left, currentSum, result);
}
if(node.right != null){
calculateSumRootToLeaf(node.right, currentSum, result);
}
}
currentSum = currentSum.substring(0, currentSum.length() - 1);
}
}

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