// 894. All Possible Full Binary Trees #include #include /* * given an integer 'n', return a list of all possible full binary trees with * 'n' nodes. each node of each tree in the answer must have 'node.val == 0'. * each element of the answer is the root node of one possible tree. you may * return the final list of trees in any order. a full binary tree is a binary * tree where each node has exactly zero or two children */ struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; }; struct TreeNode **allPossibleFBT(int n, int *return_size) { if (!n % 2) { *return_size = 0; return NULL; } if (n == 1) { struct TreeNode *root = malloc(sizeof(struct TreeNode)); root->val = 0; root->left = NULL; root->right = NULL; struct TreeNode **ans = malloc(sizeof(struct TreeNode *)); *return_size = 1; ans[0] = root; return ans; } if (n == 3) { struct TreeNode *root = malloc(sizeof(struct TreeNode)); struct TreeNode *left = malloc(sizeof(struct TreeNode)); struct TreeNode *right = malloc(sizeof(struct TreeNode)); root->val = 0; left->val = 0; right->val = 0; left->left = NULL; left->right = NULL; right->left = NULL; right->right = NULL; root->left = left; root->right = right; struct TreeNode **ans = malloc(sizeof(struct TreeNode)); *return_size = 1; ans[0] = root; return ans; } struct TreeNode **ans = malloc(5000 * sizeof(struct TreeNode)); int idx = 0; for (int i = 1; i <= (n - 1); i += 2) { int *cnt_l = calloc(1, sizeof(int)); int *cnt_r = calloc(1, sizeof(int)); struct TreeNode **l_tree = allPossibleFBT(i, cnt_l); struct TreeNode **r_tree = allPossibleFBT(n - 1 - i, cnt_r); for (int j = 0; j < (*cnt_l); j++) { for (int k = 0; k < (*cnt_r); k++) { struct TreeNode *new = malloc(sizeof(struct TreeNode)); new->val = 0; new->left = l_tree[j]; new->right = r_tree[k]; ans[idx] = new; idx++; } } free(cnt_l), free(cnt_r); } ans = realloc(ans, idx * sizeof(struct TreeNode *)); *return_size = idx; return ans; }