LeetCode #129 - Sum Root to Leaf Numbers

最新推荐文章于 2022-01-08 12:45:22 发布

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最新推荐文章于 2022-01-08 12:45:22 发布

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本文介绍了一种算法，用于求解给定二叉树中所有从根节点到叶子节点路径所代表数字的总和。通过递归遍历的方式获取所有有效路径并计算其数值总和。

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题目描述：

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.

利用递归得到从根节点到叶节点的所有路径，并计算路径之和。

class Solution {
public:
    int sumNumbers(TreeNode* root) {
        int sum=0;
        vector<int> path;
        sumNumbers_recur(root,path,sum);
        return sum;
    }
    
    void sumNumbers_recur(TreeNode* root, vector<int> path, int& sum)
    {
        if(root==NULL) return;
        else if(root->left==NULL&&root->right==NULL)
        {
            path.push_back(root->val);
            for(int i=0;i<path.size();i++) sum+=path[i]*pow(10,path.size()-1-i);
            return;
        }
        else
        {
            path.push_back(root->val);
            sumNumbers_recur(root->left,path,sum);
            sumNumbers_recur(root->right,path,sum);
        }
    }
};