【leetcode】129. Sum Root to Leaf Numbers

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本文探讨了如何在仅包含0-9数字的二叉树中找到所有从根节点到叶子节点路径所代表的数字总和。通过前序遍历算法，详细解析了递归求解过程，为读者提供了清晰的代码实现。

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.

题解如下：

采用前序遍历的方式遍历整棵树，因为由题意易知对于根结点我们要先处理。（方便递归到子结点时将之前结点的值乘以10）

/**
 * 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) {
        if(root == null) {
            return 0;
        }
        return dfs(root,0);
    }
    
    public int dfs(TreeNode root,int sum) {
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right == null) {
            return sum * 10 + root.val;
        }
        return dfs(root.left,sum * 10 + root.val) + dfs(root.right,sum * 10 + root.val);
        
    }
}