Calculate the Sum of Root To Leaf Binary Numbers in Java

The challenge

Given a binary tree, each node has value 0 or 1.  Each root-to-leaf path represents a binary number starting with the most significant bit.  For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.

Return the sum of these numbers.

Example 1:

Input: [1,0,1,0,1,0,1]
Output: 22
Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22

Note:

1. The number of nodes in the tree is between 1 and 1000.
2. node.val is 0 or 1.
3. The answer will not exceed 2^31 - 1.

Given the Binary Tree definition

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

The solution in Java

Option 1:

class Solution {
    public int sumRootToLeaf(TreeNode root) {
        int rootToLeaf = 0;
        int currNumber = 0;
        Deque<Pair<TreeNode, Integer>> stack = new ArrayDeque();
        stack.push(new Pair(root, 0));

        while (!stack.isEmpty()) {
          Pair<TreeNode, Integer> p = stack.pop();
          root = p.getKey();
          currNumber = p.getValue();

          if (root != null) {
            currNumber = (currNumber << 1) | root.val;
            if (root.left == null && root.right == null) {
              rootToLeaf += currNumber;
            } else {
              stack.push(new Pair(root.right, currNumber));
              stack.push(new Pair(root.left, currNumber));
            }
          }
        }
        return rootToLeaf;
    }
}

Option 2 (using Depth First Search):

class Solution {
    public int sumRootToLeaf(TreeNode root) {
        return dfs(root, 0);
    }
    
    private int dfs(TreeNode root, int sum) {
        if(root == null)
            return 0;
        sum = sum * 2 + root.val; // sum = (sum << 1) + root.val;
        return (root.left == null && root.right == null) ? sum : dfs(root.left, sum) + dfs(root.right, sum);
    }
}

Option 3 (using iterative):

class Solution {
    int val = 0;

    public int sumRootToLeaf(TreeNode root) {
        Solution s = new Solution();
        s.findCount(root, 0);

        return s.val;
    }

    public void findCount(TreeNode root, int value){
        if(root!=null){
            int sum = root.val + value*2;

            if(root.left == null && root.right == null){
                this.val = this.val + sum;
            }

            findCount(root.left, sum);
            findCount(root.right, sum);
        }
    }
}