You are given the root of a binary tree containing digits from 0 to 9 only.
Each root-to-leaf path in the tree represents a number.
For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.
Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.
A leaf node is a node with no children.
Example 1:
**Input:** root = [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:** root = [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.
Constraints:
The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 9
The depth of the tree will not exceed 10.
ac1: Backtracking
use StringBuilder, slower
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int sum;
public int sumNumbers(TreeNode root) {
sum = 0;
StringBuilder sb = new StringBuilder();
backtrack(root, sb);
return sum;
}
private void backtrack(TreeNode root, StringBuilder sb) {
// exit
if (root == null) return;
// leaf
sb.append(root.val);
if (root.left == null && root.right == null) {
sum += Integer.parseInt(sb.toString());
}
backtrack(root.left, sb);
backtrack(root.right, sb);
sb.setLength(sb.length() - 1);
}
}
// DFS, backtracking
ac2: devide & conquer
direct number calculation, so beautiful...
class Solution {
public int sumNumbers(TreeNode root) {
return sum(root, 0);
}
private int sum(TreeNode root, int sum) {
// exit
if (root == null) return 0;
// leaf
if (root.left == null && root.right == null) {
return sum*10 + root.val;
}
// devide & conquer
return sum(root.left, sum*10 + root.val) + sum(root.right, sum*10 + root.val);
}
}
