129. Sum Root to Leaf Numbers-优快云博客

本文探讨了如何在结点值仅包含0-9的二叉树中，通过前序遍历找出所有从根到叶子的路径所表示的整数，并计算其总和。提供了具体的算法实现和两个实例说明。

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.
Example:

Input: [1,2,3]
    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]
    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.

难度：medium

题目：给定结点值只包含0-9的二叉树，每条从根到叶子的路径表示一个整数。找出所有这样的数并返回其和。
注意：叶结点即没有左右子结点

思路：前序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        return sumNumbers(root, "");
    }
    
    private int sumNumbers(TreeNode root, String s) {
        if (null == root) {
            return 0;
        }
        s += root.val;
        if (null == root.left && null == root.right) {
            return Integer.parseInt(s);
        }
        
        return sumNumbers(root.left, s) + sumNumbers(root.right, s);
    }
}