Leetcode Sum Root to Leaf Numbers problem solution

YASH PAL, 31 July 202419 January 2026

In this Leetcode Sum Root to Leaf Numbers problem solution we have 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.

Leetcode Sum Root to Leaf Numbers problem solution

Leetcode Sum Root to Leaf Numbers problem solution in Python.

class Solution:
    def sumNumbers(self, root: TreeNode) -> int:
        s=0
        ps=""
        if root==None:
            return
        def dfs(root,ps,t):
            op=0
            
            if root==None:
                return
            if root.left==None and root.right==None:
                
                ps=ps+str(root.val)
                ps=ps[::-1]
                i=1
                op=0
                for i in range(len(ps)):
                    op=op+int(ps[i])*(10**i)
                    i=i+1
                
                t=t+op
            if root.left:
                t=dfs(root.left,ps+str(root.val),t)
            if root.right:
                t=dfs(root.right,ps+str(root.val),t)
            return t
        a=dfs(root,ps,s)
        return a

Sum Root to Leaf Numbers problem solution in Java.

class Solution {
    private int res;
    
    public int sumNumbers(TreeNode root) {
        help(root, 0);
        return res;
    }
    
    private void help(TreeNode node, int num) {
        if (node == null) {
            return;
        }
        num = num * 10 + node.val;
        if (node.left == null && node.right == null) {
            res += num;
        }
        help(node.left, num);
        help(node.right, num);
    }
}

Problem solution in C++.

long totalvalue;
    long getvalue(string s)
    {
        long r=0;
        for(int i=s.size()-1;i>=0;i--)
            r+=(s[i]-'0')*pow(10,s.size()-i-1);
        return r;
    }
    void PreoderTraversal(TreeNode* node,string curval)
    {
        if(node!=nullptr)
        {
            if(node->left!=nullptr)
                PreoderTraversal(node->left,curval+to_string(node->val));
            if(node->right!=nullptr)
                PreoderTraversal(node->right,curval+to_string(node->val));
            if(node->left==nullptr and node->right==nullptr)
               totalvalue+=getvalue(curval+to_string(node->val));
            
        }
    }
    int sumNumbers(TreeNode* root) {
        totalvalue=0;
        PreoderTraversal(root,"");
        return totalvalue;
    }

Problem solution in C.

int global_sum = 0;

void sum_root_to_leaves(struct TreeNode * root, int sum)
{
    if(root == NULL)
        return;
    
    int new_sum = (sum*10) + root->val;
    
    if((root->left == NULL) && (root->right == NULL))
    {
        global_sum += new_sum;
        return;
    }
        
    sum_root_to_leaves(root->left, new_sum);
    sum_root_to_leaves(root->right, new_sum);
   
}


int sumNumbers(struct TreeNode* root)
{
    global_sum = 0;
    int sum = 0;
    sum_root_to_leaves(root, sum);
    
    return global_sum;
    
}

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