Leetcode 129) Sum Root to Leaf Numbers

29 Sep 2021 in Algorithms

My Answer

/**
 * 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;
 *     }
 * }
 */
class Solution {
    int res=0;
    public int sumNumbers(TreeNode root) {
        helper(0 , root);
        return res;
    }

    public void helper(int prev , TreeNode tn){
        int cur=prev*10+tn.val;
        //basecase
        if(tn.left ==null && tn.right==null){
            res += cur;
            return;
        }

        if(tn.left!=null){
            //tn.left!=null
            helper(cur,tn.left);
        }
        if(tn.right!=null){
            helper(cur,tn.right);
        }
    }
}

Other Answer

class Solution {
    int rootToLeaf = 0;

    public void preorder(TreeNode r, int currNumber) {
        if (r == null) {
            return;
        }
        if (r != null) {
            currNumber = currNumber * 10 + r.val;
            // if it's a leaf, update root-to-leaf sum
            if (r.left == null && r.right == null) {
                rootToLeaf += currNumber;
                return;
            }
            preorder(r.left, currNumber);
            preorder(r.right, currNumber) ;
        }
    }

    public int sumNumbers(TreeNode root) {
        preorder(root, 0);
        return rootToLeaf;
    }
}