leetcode 103.二叉树的锯齿形层序遍历

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）。
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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;
}
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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--;
    }
}
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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
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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;
    } 
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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;
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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;
}
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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^ _ ^