129. Sum Root to Leaf Numbers

Description

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.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: [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]

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.

Solutions

TreeNode

// 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;
    }
}

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    private int sum = 0;
    
    public int sumNumbers(TreeNode root) {
        if(root == null) return sum;
        helper(root, root.val);
        return sum;
    }
    
    private void helper(TreeNode root, int pathSum) {
        if(root == null) return;
        
        if(root.left == null && root.right == null) {
            sum += pathSum;
            return;
        }

        if(root.left != null) {
            helper(root.left, pathSum*10+root.left.val);
        }
        
        if(root.right != null) {
            helper(root.right, pathSum*10+root.right.val);
        }
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        return helper(root, 0);
    }
    
    private int helper(TreeNode node, int sum) {
        if(node == null) return 0;

        sum = (sum * 10) + node.val;
        
        if(node.left == null && node.right == null) {
            return sum;
        }

        return helper(node.left, sum) + 
                helper(node.right, sum);
    }
}

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