Leetcode 1302.层数最深子叶结点的和

大家好,今天我给大家分享一下我关于这个题的想法,我这个题过程比较复杂,但大家如果觉得好的话,就请给个免费的赞吧,谢谢了^ _ ^

1.题目要求:

c 复制代码
给你一棵二叉树的根节点 root ,请你返回 层数最深的叶子节点的和 。

举例如图所示:

2.做题思路:

1.先用前序遍历求出树的结点数量:

c 复制代码
void preorder(struct TreeNode* root,int* length){
    if(root == NULL){
        return;
    }
    (*length)++;
    preorder(root->left,length);
    preorder(root->right,length);
}
int* length = (int*)malloc(sizeof(int));
    *length = 0;
    preorder(root,length);

2.然后再根据结点数量用malloc分配两个数组,一个要进行层序遍历,一个要记录树每一层的宽度:

c 复制代码
	//此数组是用来存层序遍历的
	int* treeval = (int*)malloc(sizeof(int)* (*length));
    int j = 0;
    //此数组是用来进行记录树的每层的宽度
    int* col = (int*)malloc(sizeof(int) * (*length + 1));
    int j_1 = 0;

3.然后我们进行层序遍历,设置变量把每一行结点的数量记录下来,再把每个结点存入数组中:

c 复制代码
typedef struct queue{
    struct TreeNode* data;
    struct queue* next;
}queue_t;
//入队
void insert_tail(queue_t** head,struct TreeNode* data)
{
    queue_t* newnode = (queue_t*)malloc(sizeof(queue_t));
    memset(newnode,0,sizeof(queue_t));
    newnode->data = data;
    newnode->next = NULL;
    if(*head == NULL){
        *head = newnode;
        return;
    }
    queue_t* cur = *head;
    while(cur->next != NULL){
        cur = cur->next;
    }
    cur->next = newnode;
}
//出队
struct TreeNode* pop(queue_t** head){
    if(*head == NULL){
        struct TreeNode* x = (*head)->data;
        (*head) = (*head)->next;
        return x;
    }
    struct TreeNode* x = (*head)->data;
    (*head) = (*head)->next;
    return x;
}
	//开始层序遍历
	
    int nextcount = 0;//记录树每一层的结点数量
    queue_t* quence = NULL;
    int size = 0;
    insert_tail(&quence,root);
    size++;
    while(size != 0){
        for(i = 0;i < count;i++){
            struct TreeNode* node = pop(&quence);
            treeval[j] = node->val;
            j++;
            size--;
            if(node->left != NULL){
                insert_tail(&quence,node->left);
                size++;
                nextcount++;
            }
            if(node->right != NULL){
                insert_tail(&quence,node->right);
                size++;
                nextcount++;
            }
        }
        col[j_1] = nextcount;
        j_1++;
        count = nextcount;
        nextcount = 0;
    }

3.全部代码:

c 复制代码
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
typedef struct queue{
    struct TreeNode* data;
    struct queue* next;
}queue_t;
//出队
void insert_tail(queue_t** head,struct TreeNode* data)
{
    queue_t* newnode = (queue_t*)malloc(sizeof(queue_t));
    memset(newnode,0,sizeof(queue_t));
    newnode->data = data;
    newnode->next = NULL;
    if(*head == NULL){
        *head = newnode;
        return;
    }
    queue_t* cur = *head;
    while(cur->next != NULL){
        cur = cur->next;
    }
    cur->next = newnode;
}
//入队
struct TreeNode* pop(queue_t** head){
    if(*head == NULL){
        struct TreeNode* x = (*head)->data;
        (*head) = (*head)->next;
        return x;
    }
    struct TreeNode* x = (*head)->data;
    (*head) = (*head)->next;
    return x;
}
//进行前序遍历
void preorder(struct TreeNode* root,int* length){
    if(root == NULL){
        return;
    }
    (*length)++;
    preorder(root->left,length);
    preorder(root->right,length);
}
int deepestLeavesSum(struct TreeNode* root) {
    int* length = (int*)malloc(sizeof(int));
    *length = 0;
    preorder(root,length);
    int* treeval = (int*)malloc(sizeof(int)* (*length));
    int j = 0;
    int* col = (int*)malloc(sizeof(int) * (*length + 1));
    int j_1 = 0;
    int count = 1;
    col[j_1] = count;
    j_1++;
    int i = 0;
    //记录树的每一层的宽度
    int nextcount = 0;
    queue_t* quence = NULL;
    int size = 0;
    insert_tail(&quence,root);
    size++;
    while(size != 0){
        for(i = 0;i < count;i++){
            struct TreeNode* node = pop(&quence);
            treeval[j] = node->val;
            j++;
            size--;
            if(node->left != NULL){
                insert_tail(&quence,node->left);
                size++;
                nextcount++;
            }
            if(node->right != NULL){
                insert_tail(&quence,node->right);
                size++;
                nextcount++;
            }
        }
        col[j_1] = nextcount;
        j_1++;
        count = nextcount;
        nextcount = 0;
    }
    int count1 = col[j_1 - 2];
    int sum = 0;
    int f = 0;
    for(i = j - 1;i >= 0;i--){
       sum += treeval[i];
       f++;
       if(f == count1){
            break;
       }
    }
    return sum;
}

好了,这就是我全部代码大家如果觉得好的话,就请给个免费的赞吧,谢谢了^ _ ^ .

相关推荐
MicroTech202517 小时前
MLGO微算法科技通过 Lindbladians 设计线性微分方程的近似最优量子算法——开放量子系统框架下的量子ODE求解新范式
科技·算法·量子计算
知乎的哥廷根数学学派17 小时前
基于多尺度特征提取和注意力自适应动态路由胶囊网络的工业轴承故障诊断算法(Pytorch)
开发语言·网络·人工智能·pytorch·python·算法·机器学习
源代码•宸17 小时前
Leetcode—85. 最大矩形【困难】
经验分享·算法·leetcode·职场和发展·golang·单调栈
平哥努力学习ing18 小时前
《数据结构》-第八章 排序
数据结构·算法·排序算法
CoovallyAIHub18 小时前
为AI装上“纠偏”思维链,开源框架Robust-R1显著提升多模态大模型抗退化能力
深度学习·算法·计算机视觉
小棠师姐18 小时前
随机森林原理与实战:如何解决过拟合问题?
算法·机器学习·随机森林算法·python实战·过拟合解决
范纹杉想快点毕业18 小时前
欧几里得算法与扩展欧几里得算法,C语言编程实现(零基础全解析)
运维·c语言·单片机·嵌入式硬件·算法
f***241118 小时前
Bug悬案:技术侦探的破案指南
算法·bug
Swift社区18 小时前
LeetCode 472 连接词
算法·leetcode·职场和发展
Dream it possible!18 小时前
LeetCode 面试经典 150_二分查找_搜索旋转排序数组(114_33_C++_中等)
c++·leetcode·面试