1022.从根到叶的二进制之和

目录

一、题目

二、代码

---

一、题目

二、代码

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:

    int PreOrder(TreeNode*root,int sum)
    {
        if(root == nullptr)
        {
            return 0;
        }

        sum=sum<<1 |root->val;

        if(root->left==nullptr && root->right==nullptr)
        {
            return sum;
        }

        return PreOrder(root->left,sum)+PreOrder(root->right,sum);
    }
    int sumRootToLeaf(TreeNode* root) {
        int sum=0;
       return PreOrder(root,sum);
    }
};