129. Sum Root to Leaf Numbers

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

An example is the root-to-leaf path 1->2->3which 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;
    }
}