1022. 从根到叶的二进制数之和

 

/**
 * 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 sum = 0;
    int sumRootToLeaf(TreeNode* root) {
        preOrder(root, 0); 
        return sum;
    }

    void preOrder(TreeNode* node, int k) {
        if(node == nullptr) return ;
        k = (k << 1) + node->val;
        if(node->left == nullptr && node->right == nullptr) {
            sum += k;
        }
        preOrder(node->left, k);
        preOrder(node->right, k);
    }
};

  

原文地址:https://www.cnblogs.com/mjn1/p/14283283.html