LeetCode 之 Sum Root to Leaf Numbers

最新推荐文章于 2025-11-29 18:57:13 发布

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#BST #DFS #leetcode

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.

属于二叉树遍历的问题，只不过我们要存储每一个从根到叶的数值，可以用深度优先的原则，此外一个变量存储总和，一个变量存储每个路径的数值，代码如下：

/**
 * 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) {
        int sum=0;
        int currentSum=0;
        if(!root) return 0;
        findSum(root,sum,currentSum);
        return sum;
    }
    void findSum(TreeNode* root,int &sum,int currentSum){
        currentSum=currentSum*10+root->val;
        if(!root->left&&!root->right)
            sum+=currentSum;
        if(root->left)
            findSum(root->left,sum,currentSum);
        if(root->right)
            findSum(root->right,sum,currentSum);
    }
};

要注意sum应该是引用参数，这样在调用findSum函数的时候不会变成值传递。