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:

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:
在这里插入图片描述
2.做题思路:
1.先用前序遍历求出树的结点数量:

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:

	//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:

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:

/*** 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;
}