1.题目要求:
给你二叉树的根节点 root ,返回其节点值的 锯齿形层序遍历 。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
2.做题思路:由题我们可以判断,树中每到偶数层时,就会逆置,所以我们可以采用层序遍历和逆置函数的方法来解此题
3.做题步骤:
1.我们先把层序遍历所需要的队列结构,出队和入队函数写好:
//创建队列
typedef struct queue{
struct TreeNode* value;
struct queue* next;
}queue_t;
//入队
void push(queue_t** head,struct TreeNode* data){
queue_t* newnode = (queue_t*)malloc(sizeof(queue_t));
newnode->value = data;
newnode->next = NULL;
if(*head == NULL){
*head = newnode;
return;
}
queue_t* tail = *head;
while(tail->next != NULL){
tail = tail->next;
}
tail->next = newnode;
}
//出队
struct TreeNode* pop(queue_t** head){
struct TreeNode* x = (*head)->value;
(*head) = (*head)->next;
return x;
}
2.写好逆置函数:
void reverse(int* number,int left,int right){
while(left <= right){
int temp = number[left];
number[left] = number[right];
number[right] = temp;
left++;
right--;
}
}
3.写好进行层序遍历的变量:
*returnSize = 0;
if(root == NULL){
return NULL;
}
int* each_line_nodes = (int*)malloc(sizeof(int)*2000);//记录每行结点数
int j_1 = 0;
int* level_order_number = (int*)malloc(sizeof(int)* 2000);//层序遍历的数组
int j_2 = 0;
int depth = 0;//树的高度
int count = 1;//根结点的个数
int index = 0;//记录每层的第一个的索引
int nextcount = 0;//下一个结点的个数
int size = 0;//记录队列中的个数
queue_t* quence = NULL;//设置队列
4.进行层序遍历:
//进行层序遍历
push(&quence,root);
size++;
while(size != 0){
depth++;
for(int i = 0;i < count;i++){
struct TreeNode* temp = pop(&quence);
size--;
level_order_number[j_2] = temp->val;
j_2++;
if(temp->left != NULL){
push(&quence,temp->left);
size++;
nextcount++;
}
if(temp->right != NULL){
push(&quence,temp->right);
size++;
nextcount++;
}
}
each_line_nodes[j_1] = count;
j_1++;
//如果高度为偶数时,就要对数组这一次的元素进行逆置
if(depth % 2 == 0){
reverse(level_order_number,index,j_2 - 1);
index += count;
}else{
index += count;
}
count = nextcount;
nextcount = 0;
}
5.再创造二维数组,把值放入二维数组中:
//设立二维数组
int** array = (int**)malloc(sizeof(int*)* depth);
for(int i = 0;i < depth;i++){
array[i] = (int*)malloc(sizeof(int) * each_line_nodes[i]);
}
int f = 0;
for(int i = 0;i < depth;i++){
for(int j = 0;j < each_line_nodes[i];j++){
array[i][j] = level_order_number[f];
f++;
}
}
*returnSize = depth;
*returnColumnSizes = (int*)malloc(sizeof(int) * (*returnSize));
for(int i = 0;i < depth;i++){
(*returnColumnSizes)[i] = each_line_nodes[i];
}
return array;
以下为全部代码:
/**
* 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().
*/
//创建队列
typedef struct queue{
struct TreeNode* value;
struct queue* next;
}queue_t;
//入队
void push(queue_t** head,struct TreeNode* data){
queue_t* newnode = (queue_t*)malloc(sizeof(queue_t));
newnode->value = data;
newnode->next = NULL;
if(*head == NULL){
*head = newnode;
return;
}
queue_t* tail = *head;
while(tail->next != NULL){
tail = tail->next;
}
tail->next = newnode;
}
//出队
struct TreeNode* pop(queue_t** head){
struct TreeNode* x = (*head)->value;
(*head) = (*head)->next;
return x;
}
void reverse(int* number,int left,int right){
while(left <= right){
int temp = number[left];
number[left] = number[right];
number[right] = temp;
left++;
right--;
}
}
int** zigzagLevelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
*returnSize = 0;
if(root == NULL){
return NULL;
}
int* each_line_nodes = (int*)malloc(sizeof(int)*2000);//记录每行结点数
int j_1 = 0;
int* level_order_number = (int*)malloc(sizeof(int)* 2000);//层序遍历的数组
int j_2 = 0;
int depth = 0;//树的高度
int count = 1;//根结点的个数
int index = 0;//记录每层的第一个的索引
int nextcount = 0;//下一个结点的个数
int size = 0;//记录队列中的个数
queue_t* quence = NULL;//设置队列
//进行层序遍历
push(&quence,root);
size++;
while(size != 0){
depth++;
for(int i = 0;i < count;i++){
struct TreeNode* temp = pop(&quence);
size--;
level_order_number[j_2] = temp->val;
j_2++;
if(temp->left != NULL){
push(&quence,temp->left);
size++;
nextcount++;
}
if(temp->right != NULL){
push(&quence,temp->right);
size++;
nextcount++;
}
}
each_line_nodes[j_1] = count;
j_1++;
//如果高度为偶数时,就要对数组这一次的元素进行逆置
if(depth % 2 == 0){
reverse(level_order_number,index,j_2 - 1);
index += count;
}else{
index += count;
}
count = nextcount;
nextcount = 0;
}
//设立二维数组
int** array = (int**)malloc(sizeof(int*)* depth);
for(int i = 0;i < depth;i++){
array[i] = (int*)malloc(sizeof(int) * each_line_nodes[i]);
}
int f = 0;
for(int i = 0;i < depth;i++){
for(int j = 0;j < each_line_nodes[i];j++){
array[i][j] = level_order_number[f];
f++;
}
}
*returnSize = depth;
*returnColumnSizes = (int*)malloc(sizeof(int) * (*returnSize));
for(int i = 0;i < depth;i++){
(*returnColumnSizes)[i] = each_line_nodes[i];
}
return array;
}
好了,这就是我的解题方法,大家如果觉得好的话,不妨给个免费的赞吧,谢谢了^ _ ^