C++ | Level-Order Traversal Approach |
Intuition\n Describe your first thoughts on how to solve this problem. The given code is a C++ implementation of a function maxLevelSum that aims to find the le
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->The given code is a C++ implementation of a function `maxLevelSum` that aims to find the level in a binary tree that has the maximum sum of node values. Here are my initial thoughts on how the code works and how to solve the problem:\n\n1. The function `maxLevelSum` takes a pointer to the root node of the binary tree as input and returns an integer representing the level with the maximum sum.\n\n2. The code initializes variables `sum` and `ans` to store the maximum sum and the corresponding level, respectively. The variable `l` is used to keep track of the current level.\n\n3. It uses a queue data structure to perform a level-order traversal of the binary tree. The root node is enqueued initially.\n\n4. The while loop iterates until the queue is empty, indicating that all levels of the binary tree have been processed.\n\n5. Inside the while loop, it retrieves the number of nodes in the current level by getting the size of the queue. It also initializes a variable `s` to store the sum of node values for the current level.\n\n6. A nested while loop iterates `n` times, where `n` is the number of nodes in the current level. It dequeues a node from the front of the queue, adds its value to `s`, and enqueues its left and right child nodes if they exist.\n\n7. After processing all nodes in the current level, it compares the sum `s` with the current maximum sum `sum`. If `s` is greater, it updates `sum` and assigns the current level `l` to `ans`.\n\n8. Finally, it increments `l` to move to the next level.\n\n9. The function returns `ans` after processing all levels of the binary tree.\n\nTo solve the problem, the code uses a level-order traversal to visit each level and calculate the sum of node values. It keeps track of the level with the maximum sum and returns that level as the result.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->In summary, the approach taken by the code is to perform a level-order traversal of the binary tree and calculate the sum of node values for each level. It keeps track of the maximum sum and the corresponding level as it traverses the tree, and returns the level with the maximum sum at the end.\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     TreeNode *left;\n *     TreeNode *right;\n *     TreeNode() : val(0), left(nullptr), right(nullptr) {}\n *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n    int maxLevelSum(TreeNode* root)\n     {\n        int sum = INT_MIN, ans, l = 1;\n        queue<TreeNode*> q;\n        q.push(root);\n        while(q.size())\n         {\n            int n = q.size(), s = 0;\n            while(n--)\n             {\n                TreeNode* temp = q.front(); q.pop();\n                s += temp -> val;\n                if(temp -> left) q.push(temp -> left);\n                if(temp -> right) q.push(temp -> right);\n            }\n            if(s > sum) sum = s, ans = l;\n            l++;\n        }\n        return ans;\n    }\n};\n```\n![Leetcode.jpeg](https://assets.leetcode.com/users/images/e3c6a816-9433-4827-a216-35c8680bfea7_1686790302.861607.jpeg)\n
['Depth-First Search', 'Breadth-First Search', 'Queue', 'Binary Tree', 'C++']