117. Sum Root to Leaf Numbers

Carry the number formed so far down each path. At each node the running number becomes cur*10 + node.val. When you reach a leaf, that running value is a complete root-to-leaf number; sum those over all leaves.

Idea

Each root-to-leaf path spells a decimal number: descending one level appends a digit, i.e. multiply the number so far by 10 and add the new node's value. A DFS that threads this cur value downward hits every leaf with its finished number; summing those gives the answer.

Algorithm

dfs(node, cur):

1. If node is null, return 0.
2. cur = cur * 10 + node.val.
3. If node is a leaf, return cur.
4. Otherwise return dfs(left, cur) + dfs(right, cur).

Start with dfs(root, 0).

Worked example: [4,9,0,5,1]

    4
   / \
  9   0
 / \
5   1

Paths: 4→9→5 = 495, 4→9→1 = 491, 4→0 = 40. Sum = 1026.

Complexity

• Time: O(n).
• Space: O(h) recursion depth.

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

    private int dfs(TreeNode node, int cur) {
        if (node == null) {
            return 0;
        }
        cur = cur * 10 + node.val;
        if (node.left == null && node.right == null) {
            return cur;
        }
        return dfs(node.left, cur) + dfs(node.right, cur);
    }
}