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// 1161. Maximum Level Sum of a Binary Tree
#include <limits.h>
#include <stdlib.h>
/*
* given the 'root' of a binary tree, the level of its root is 1, the level of
* its children is 2, and so on. return the smallest level 'x' such that the sum
* of all the values of nodes at level 'x' is maximal.
*/
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
};
int maxLevelSum(struct TreeNode *root) {
struct TreeNode **even = malloc(5000 * sizeof(struct TreeNode *));
struct TreeNode **odd = malloc(5000 * sizeof(struct TreeNode *));
int even_idx = 0, odd_idx = 0;
even[0] = root, even_idx = 1;
int max_sum = INT_MIN, max_id = -1, level = 1;
while (even_idx || odd_idx) {
int sum = 0;
if (even_idx) {
for (int i = 0; i < even_idx; i++) {
sum += even[i]->val;
if (even[i]->left) {
odd[odd_idx] = even[i]->left;
odd_idx++;
}
if (even[i]->right) {
odd[odd_idx] = even[i]->right;
odd_idx++;
}
}
if (sum > max_sum) {
max_sum = sum;
max_id = level;
}
even_idx = 0;
level++;
} else {
for (int i = 0; i < odd_idx; i++) {
sum += odd[i]->val;
if (odd[i]->left) {
even[even_idx] = odd[i]->left;
even_idx++;
}
if (odd[i]->right) {
even[even_idx] = odd[i]->right;
even_idx++;
}
}
if (sum > max_sum) {
max_sum = sum;
max_id = level;
}
odd_idx = 0;
level++;
}
}
return max_id;
}
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