Sum Root to Leaf Numbers

Leet code link - https://leetcode.com/problems/sum-root-to-leaf-numbers/description/

Complexity - Medium

Description

You are given the root of a binary tree containing digits from 0 to 9 only.

Each root-to-leaf path in the tree represents a number.

For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123. Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

A leaf node is a node with no children.

Example 1:

Input: root = [1,2,3]
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: root = [4,9,0,5,1]
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

Constraints:

The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 9
The depth of the tree will not exceed 10.

Solution

To solve the problem of finding the sum of numbers represented by root-to-leaf paths in a binary tree using a Depth-First Search (DFS) approach, we can follow these steps:

Use DFS Algorithm:

Implement a DFS traversal of the binary tree, starting from the root. Maintain a variable to keep track of the current number formed from the root to the current node as you traverse. DFS Recursive Function:

Define a recursive function (dfs function) that takes the current node, the current path value, and a variable to accumulate the sum of all root-to-leaf path numbers. If the current node is None, return immediately (base case for leaf nodes). Update the current path value by appending the current node's value (e.g., path_sum = path_sum * 10 + node.val). If the current node is a leaf node (i.e., it has no children), add the path_sum to the sum. Recursively call the function for the left and right children of the current node, passing the updated path_sum. Initialize and Call the DFS Function:

Start the DFS traversal from the root node with an initial path of 0. Accumulate the sum of all root-to-leaf path numbers using the DFS function. Return the Result:

After the DFS traversal completes, return the accumulated sum, which represents the total sum of all root-to-leaf path numbers in the binary tree.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int dfs(TreeNode root, int pathSum){
        if(root == null) return 0;
        pathSum = pathSum*10 + root.val;
        if(root.left == null && root.right == null) return pathSum;
        return dfs(root.left,pathSum) + dfs(root.right,pathSum);

    }
    public int sumNumbers(TreeNode root) {
        return dfs(root,0);
    }
}

Complexity

Time Complexity - O(N)

Space Complexity - O(N)