Sum Root to Leaf Numbers

最新推荐文章于 2026-01-05 22:36:07 发布

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本文介绍了一种算法，用于求解二叉树中所有从根节点到叶子节点的路径数值之和。通过递归遍历的方式计算每条路径所代表的数字，并最终累加得到总和。

题目：

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 number123.

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.

思路：

类似于前面的一篇博文 Path Sum

代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<int> result;
public:
    int sumNumbers(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(!root)
            return 0;
        int sum = 0;
        result.clear();
        pathNum(root,sum);
        sum=0;
        for(int i=0;i<result.size();i++)
            sum+=result[i];
        return sum;
        
    }
    void pathNum(TreeNode *root, int cursum)
    {
        if(!root)
            return;
        cursum = 10*cursum+root->val;
        if(root->left==NULL && root->right==NULL)
        {
            result.push_back(cursum);
            return;
        }
        if(root->left!=NULL)
            pathNum(root->left,cursum);
        if(root->right!=NULL)
            pathNum(root->right,cursum);   
        return;   
    }    
};