medium Sum of Path Numbers

Problem Statement

Given a binary tree where each node can only have a digit (0-9) value, each root-to-leaf path will represent a number. Find the total sum of all the numbers represented by all paths.

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.

Try it yourself

Try solving this question here:

🎯 STRICT STANDOUT — Sum of Path Numbers

1. Why

Root→leaf as decimals; num=num*10+val; sum at leaves.

2. K11

Θ(n)/O(h). long if needed.

3. K12

DECIMAL PATH FOLD. When-NOT: binary *2; path sum target.

4. K13

single digit; multi-digit paths.

5. Panel

Q1. Formula? A: 10*num+val. Q2. Time Θ(n). Q3. Space O(h).

✅ Solution Sum of Path Numbers

Problem Statement

Given a binary tree where each node can only have a digit (0-9) value, each root-to-leaf path will represent a number. Find the total sum of all the numbers represented by all paths.

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

This problem follows the Binary Tree Path Sum pattern. We can follow the same DFS approach. The additional thing we need to do is to keep track of the number representing the current path.

How do we calculate the path number for a node? Taking the first example mentioned above, say we are at node ‘7’. As we know, the path number for this node is ‘17’, which was calculated by: 1 * 10 + 7 => 17. We will follow the same approach to calculate the path number of each node.

Here is the visual representation of the algorithm:

Code

Here is the code for this algorithm:

java

// class TreeNode {
//   int val;
//   TreeNode left;
//   TreeNode right;

//   TreeNode(int x) {
//     val = x;
//   }
// };

class Solution {

  public int findSumOfPathNumbers(TreeNode root) {
    return findRootToLeafPathNumbers(root, 0);
  }

  private static int findRootToLeafPathNumbers(
    TreeNode currentNode,
    int pathSum
  ) {
    if (currentNode == null) return 0;

    // calculate the path number of the current node
    pathSum = 10 * pathSum + currentNode.val;

    // if the current node is a leaf, return the current path sum.
    if (currentNode.left == null && currentNode.right == null) {
      return pathSum;
    }

    // traverse the left and the right sub-tree
    return (
      findRootToLeafPathNumbers(currentNode.left, pathSum) +
      findRootToLeafPathNumbers(currentNode.right, pathSum)
    );
  }

  public static void main(String[] args) {
    Solution sol = new Solution();
    TreeNode root = new TreeNode(1);
    root.left = new TreeNode(0);
    root.right = new TreeNode(1);
    root.left.left = new TreeNode(1);
    root.right.left = new TreeNode(6);
    root.right.right = new TreeNode(5);
    System.out.println(
      "Total Sum of Path Numbers: " + sol.findSumOfPathNumbers(root)
    );
  }
}

Time Complexity

The time complexity of the above algorithm is O(N), where ‘N’ is the total number of nodes in the tree. This is due to the fact that we traverse each node once.

Space Complexity

The space complexity of the above algorithm will be O(N) in the worst case. This space will be used to store the recursion stack. The worst case will happen when the given tree is a linked list (i.e., every node has only one child).

🎯 STRICT STANDOUT — Solution Sum of Path Numbers

1. Why

DFS carry num; leaf ans+=num.

2. K11

Θ(n)/O(h). Example path 12+13 style offline.

3. K12

*10 FOLD. When-NOT: string parse each path.

4. K13

null 0; single.

5. Panel

Q1. Overflow? A: long. Q2. Time Θ(n). Q3. Space O(h).

🧩 Pattern · Trees

Recognize it: Hierarchical data → recurse on children; BFS (queue) for level-order, DFS (recursion) for paths.

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🔁 Drill the pattern

Lock in recognition with look-alikes.

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