LeetCode 129 求根节点到叶节点数字之和

  • 分析
    什么递归都可以,traverse函数中number是从树根节点到root节点的父节点的数字之和,sum即为所求。利用遍历树的框架锻炼递归思维。
  • 代码
/**
 * 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;
        traverse(root, 0, sum);
        return sum;
    }

    void traverse(TreeNode* root, int number, int& sum){
    
    
        if(root == nullptr) return;
        number = number * 10 + root -> val;
        
        traverse(root -> left, number, sum);
        if(root -> left == nullptr &&  root -> right == nullptr){
    
    
            sum += number;
        }
        traverse(root -> right, number, sum);
    }
};

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转载自blog.csdn.net/xiaoan08133192/article/details/115385094