## 129. Sum Root to Leaf Numbers

Given a binary tree containing digits from `0-9`only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path `1->2->3`which represents the number `123`.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

``````Input: [1,2,3]
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: [4,9,0,5,1]
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.
``````

Thoughts:

1. Recusively: Having a current sum as parameter add detect whether current node is leaf node, if it is adds the sum to the total sum
2. Iteratively: Using Stack

Code: Recursively:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
int sum;
public int sumNumbers(TreeNode root) {
sum = 0;
if(root != null) sumNumbers(root, 0);
return sum;
}
private void sumNumbers(TreeNode root, int cur){
cur = cur * 10 + root.val;
// leaf node
if(root.left == null && root.right == null) {
sum += cur;
return;
}

if(root.left != null) sumNumbers(root.left,cur);
if(root.right != null) sumNumbers(root.right,cur);
}

}
``````

Code: Iteratively

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
if (root == null) return 0;

Stack<TreeNode> s = new Stack<>();
s.push(root); int sum = 0;
while(!s.empty()){
TreeNode cur = s.pop();
if(cur.left == null && cur.right == null){
sum += cur.val;
continue;
}
if(cur.left != null){
cur.left.val += cur.val *10;
s.push(cur.left);
}
if(cur.right != null){
cur.right.val += cur.val* 10;
s.push(cur.right);
}
}

return sum;
}
}
``````