Sum Root to Leaf Numbers

 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.


 Approach: DFS (Depth First Search)

Steps:

  1. Start from the root.

  2. Pass the current number so far down the path.

  3. At each node: current = current * 10 + node.val

  4. When you reach a leaf, return the number.

  5. Recursively calculate left and right subtree sum


/**
 * 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 sumNumbers(TreeNode root) {
        return dfs(root, 0);
    }

    // Helper function
    private int dfs(TreeNode node, int currentSum) {
        if (node == null) return 0;

        // Update the path number
        currentSum = currentSum * 10 + node.val;

        // If it's a leaf node, return the final number
        if (node.left == null && node.right == null) {
            return currentSum;
        }

        // Recur for left and right subtrees and return the total sum
        return dfs(node.left, currentSum) + dfs(node.right, currentSum);
    }
}

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