C语言刷二叉树(二)基础部分
2024-08-28 00:55:41
102. 二叉树的层序遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/ /**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes)
{
struct TreeNode *queue[2001];
int front = 0, rear = 0; if (root != NULL) {
queue[rear++] = root;
} int **res = (int **)malloc(sizeof(int *) * 2001);
*returnColumnSizes = (int *)malloc(sizeof(int) * 2001);
*returnSize = 0; while (front < rear) {
int colIndex = 0;
// 申请内存
int last = rear;
res[(*returnSize)] = (int *)malloc(sizeof(int) * (last - front));
while (front < last) {
// 取出当前节点
struct TreeNode *curNode = queue[front++];
res[(*returnSize)][colIndex++] = curNode->val; if (curNode->left != NULL) {
queue[rear++] = curNode->left;
}
if (curNode->right != NULL) {
queue[rear++] = curNode->right;
}
}
(*returnColumnSizes)[(*returnSize)] = colIndex;
(*returnSize)++;
}
return res;
}
226. 翻转二叉树
这道题注意:写指针的Swap的时候,要取指针的地址,即二级指针
方法1:递归(前序遍历)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/ void Swap(struct TreeNode **left, struct TreeNode **right)
{
// printf("begin left:%p, right:%p\n", left, right);
struct TreeNode *tmp = *left;
*left = *right;
*right = tmp;
// printf("end left:%p, right:%p\n", left, right);
} struct TreeNode* invertTree(struct TreeNode* root)
{
if (root == NULL) {
return root;
}
// 前序:中 左 右
// printf("Swap前 end left:%p, right:%p\n", root->left, root->right);
Swap(&root->left, &root->right);
// printf("Swap后 end left:%p, right:%p\n", root->left, root->right);
// struct TreeNode* temp = root->left;
// root->left = root->right;
// root->right = temp; invertTree(root->left);
invertTree(root->right);
return root;
}
方法2:迭代(前序遍历)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/ void Swap(struct TreeNode **left, struct TreeNode **right)
{
struct TreeNode *tmp = *left;
*left = *right;
*right = tmp;
} struct TreeNode* invertTree(struct TreeNode* root)
{
struct TreeNode *stack[10];
int index = 0; if (root != NULL) {
stack[index++] = root;
}
while (index != 0) {
// 取出当前节点(准备交换左右子节点)
struct TreeNode *curNode = stack[--index];
if (curNode == NULL) {
break;
}
Swap(&curNode->left, &curNode->right);
// 上面已经交换完了,这里安装前序遍历正常顺序(如果不是翻转的需要则逆序,由于是栈)
if (curNode->left != NULL) {
stack[index++] = curNode->left;
}
if (curNode->right != NULL) {
stack[index++] = curNode->right;
}
}
return root;
}
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