[leetcode] 129. Sum Root to Leaf Numbers

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本文介绍了一种解决二叉树中所有从根节点到叶子节点路径数字之和的方法。通过深度优先搜索（DFS）递归算法实现，针对每个节点计算路径数值并累加求和。

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.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

解法一：

recursion的解法，对应dfs。对每一个node来说，如果它是left, 应该返回sum*10+val，sum是之前拿到的值。不然，继续迭代。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        return sumNumbersDFS(root, 0);
    }
    
    int sumNumbersDFS(TreeNode* root, int sum){
        if(!root) return 0;
        sum = sum*10 + root->val;
        if(!root->left && !root->right) return sum;
        return sumNumbersDFS(root->left, sum) + sumNumbersDFS(root->right,sum);
    }
};