DFS Sum Root to Leaf Numbers

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思路：

DFS。

时间复杂度O(N)，空间复杂度O(logN)。

/** * 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 {private:    void dfs(int sum, int &total_sum, TreeNode* root) {        if(root == NULL) {            return;        }        if(root->left == NULL && root->right == NULL) {            sum = sum*10 + root->val;            total_sum += sum;            return;        }        dfs(sum*10+root->val, total_sum, root->left);        dfs(sum*10+root->val, total_sum, root->right);    }public:    int sumNumbers(TreeNode* root) {        int sum = 0;        int total_sum = 0;        dfs(sum, total_sum, root);        return total_sum;    }};

优化，去掉sum和total_sum中间变量。

/** * 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 {private:    int dfs(TreeNode* root, int sum) {        if(root == NULL) return 0;        if(root->left == NULL && root->right == NULL) {            return 10*sum + root->val;        }        return dfs(root->left, 10*sum + root->val) + dfs(root->right, 10*sum + root->val);    }public:    int sumNumbers(TreeNode* root) {        return dfs(root, 0);    }};

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