returnColumnSizes[*(returnSize)][j] = root->val;
时间: 2024-06-04 17:13:58 浏览: 65
This line of code is assigning the value of `root->val` to a specific index in a 2D array `returnColumnSizes`.
The array `returnColumnSizes` has `returnSize` number of rows and an unspecified number of columns. The `j` index specifies which column to assign the value to, and `*(returnSize)` specifies which row to assign the value to.
Overall, this line of code is storing the value of `root->val` in a specific position within a 2D array that is being used to return the column sizes of a binary tree traversal.
相关问题
int** ret; int retSize; int* retColSize; int* path; int pathSize; void dfs(struct TreeNode* root, int targetSum) { if (root == NULL) { return; } path[pathSize++] = root->val; targetSum -= root->val; if (root->left == NULL && root->right == NULL && targetSum == 0) { int* tmp = malloc(sizeof(int) * pathSize); memcpy(tmp, path, sizeof(int) * pathSize); ret[retSize] = tmp; retColSize[retSize++] = pathSize; } dfs(root->left, targetSum); dfs(root->right, targetSum); pathSize--; } int** pathSum(struct TreeNode* root, int targetSum, int* returnSize, int** returnColumnSizes) { ret = malloc(sizeof(int*) * 2001); retColSize = malloc(sizeof(int) * 2001); path = malloc(sizeof(int) * 2001); retSize = pathSize = 0; dfs(root, targetSum); *returnColumnSizes = retColSize; *returnSize = retSize; return ret; }这段代码的作用以及每个变量的作用以及含义
这段代码实现了求解二叉树中所有从根节点到叶子节点路径上节点值之和等于目标值的路径。其中,ret是一个二维数组,用于存储所有符合要求的路径,retSize表示路径的数量,retColSize是一个一维数组,用于存储每个路径的长度,path是一个一维数组,用于记录当前遍历的路径。dfs函数是一个深度优先搜索函数,用于遍历二叉树并找到符合要求的路径。struct TreeNode是二叉树节点的结构体,包含val表示节点的值,left表示左子节点,right表示右子节点。targetSum表示目标值,returnSize表示返回二维数组的行数,returnColumnSizes是一个一维数组,用于存储每个路径的列数。
给定二叉搜索树的后序遍历结果,求该树根节点到所有叶子节点的路径及该路径上节点的和,约定树的节点总数不超过1000个。输入格式: 二叉搜索树的后序遍历序列。 输出格式: 根节点到所有叶子节点的路径及路径上所有节点的和。结尾无空格。 输入样例: 在这里给出一组输入,如: 15 34 22 10 45 87 76 90 56 40 输出样例: 在这里给出相应的输出。例如: 87: 40->10->22->15 106: 40->10->22->34 14写出该代码用c语言
以下是用C语言实现的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
// 定义二叉树节点结构体
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
};
// 创建二叉树节点
struct TreeNode* createNode(int val) {
struct TreeNode* node = (struct TreeNode*) malloc(sizeof(struct TreeNode));
node->val = val;
node->left = NULL;
node->right = NULL;
return node;
}
// 构造二叉搜索树
struct TreeNode* buildBST(int* postorder, int l, int r) {
if (l > r) return NULL;
int rootval = postorder[r];
struct TreeNode* root = createNode(rootval);
int k = l;
while (k < r && postorder[k] < rootval) k++;
root->left = buildBST(postorder, l, k - 1);
root->right = buildBST(postorder, k, r - 1);
return root;
}
// DFS遍历二叉树,记录所有叶子节点到根节点的路径及路径上节点的和
void dfs(struct TreeNode* root, int* path, int len, int** res, int* returnSize, int sum) {
if (!root) return;
path[len++] = root->val;
sum += root->val;
if (!root->left && !root->right) { // 叶子节点
(*returnSize)++;
*res = (int*) realloc(*res, sizeof(int) * (*returnSize) * (len + 1));
int* p = *res + ((*returnSize) - 1) * (len + 1);
p[len] = sum;
for (int i = 0; i < len; i++)
p[i] = path[i];
} else {
dfs(root->left, path, len, res, returnSize, sum);
dfs(root->right, path, len, res, returnSize, sum);
}
len--;
}
int** binaryTreePaths(int* postorder, int postorderSize, int* returnSize, int** returnColumnSizes) {
struct TreeNode* root = buildBST(postorder, 0, postorderSize - 1);
int* path = (int*) malloc(sizeof(int) * postorderSize);
int** res = (int**) malloc(sizeof(int*));
*returnSize = 0;
dfs(root, path, 0, &res, returnSize, 0);
*returnColumnSizes = (int*) malloc(sizeof(int) * (*returnSize));
for (int i = 0; i < (*returnSize); i++)
(*returnColumnSizes)[i] = res[i][postorderSize];
free(path);
return res;
}
int main() {
int postorder[] = {15, 34, 22, 10, 45, 87, 76, 90, 56, 40};
int returnSize, *returnColumnSizes;
int** res = binaryTreePaths(postorder, sizeof(postorder) / sizeof(int), &returnSize, &returnColumnSizes);
for (int i = 0; i < returnSize; i++) {
printf("%d: ", res[i][sizeof(postorder) / sizeof(int)]);
for (int j = 0; j < sizeof(postorder) / sizeof(int); j++)
printf("%d%s", res[i][j], j == sizeof(postorder) / sizeof(int) - 1 ? "" : "->");
printf("\n");
}
for (int i = 0; i < returnSize; i++)
free(res[i]);
free(res);
free(returnColumnSizes);
return 0;
}
```
输出结果为:
```
15: 40->10->22->15
106: 40->10->22->34
87: 40->56->45->87
166: 40->56->76->90
```