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
Examples
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.
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
// 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;
}
}
/**
* 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);
}
}
}
/**
* 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);
}
}
Follow up
