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【无标题】129. 求根节点到叶节点数字之和

我阿霆哥 2022-01-16 阅读 54
/**
 * 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:
    vector<int> ans;
    vector<int> temp;
    int sumNumbers(TreeNode* root) {
        int res=0;
        if(!root){
            return 0;
        }
        dfs(root);
        for(auto x:ans){
            res+=x;
        }
        return res;
    }

    int vecToInt(vector<int>& vec){
        int n=vec.size();
        int res=0;
       for(int i=0;i<n;++i){
           res+=(pow(10,n-i-1)*vec[i]);
       }
       return res;
    }

    void dfs(TreeNode* root){
        if(!root){
            return;
        }

        temp.push_back(root->val);//这里面要先存数据,

        if(!root->left && !root->right){
            ans.push_back(vecToInt(temp));
        }

        
        if(root->left){
            dfs(root->left);
        }
        if(root->right){
            dfs(root->right);
        }
        temp.pop_back();
    }
};
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