//题目:https://leetcode.cn/problems/binary-tree-inorder-traversal/
//方法一:递归
//思路与算法
//
//首先我们需要了解什么是二叉树的中序遍历:
// 按照访问左子树——根节点——右子树的方式遍历这棵树,
// 而在访问左子树或者右子树的时候我们按照同样的方式遍历,
// 直到遍历完整棵树。因此整个遍历过程天然具有递归的性质,
// 我们可以直接用递归函数来模拟这一过程。
//
//定义 inorder(root) 表示当前遍历到 root 节点的答案,那么按照定义,
// 我们只要递归调用 inorder(root.left) 来遍历 root 节点的左子树,
// 然后将 root 节点的值加入答案,再递归调用inorder(root.right)
// 来遍历 root 节点的右子树即可,递归终止的条件为碰到空节点。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct TreeNode {
int val;
struct TreeNode* left = NULL;
struct TreeNode* right = NULL;
}TreeNode;
//反序列化:https://leetcode.cn/problems/serialize-and-deserialize-binary-tree/solution/c-bfs-ceng-xu-bian-li-by-hukp-kare/
TreeNode* deserialize(char* data)
{
#define N 20002 // less than 20002 would break some test case.
if (!data) {
return NULL;
}
struct TreeNode* root = (struct TreeNode*) malloc(sizeof(struct TreeNode) * N);
struct TreeNode* node;
struct TreeNode *queue[N]; //通过队列,完成不断更新root的作用
int front = 0;
int rear = 0;
char* token = strtok(data, ",");
root->val = atoi(token);
queue[rear++] = root;
while (token != NULL) {
node = queue[front++];
token = strtok(NULL, ","); //左子树
if (!token) { //最后一次调用strtok,将会返回NULL,因此,这里用NULL作为while终止条件
break;
}
if (token[0] != '#') {
struct TreeNode *node_l = (struct TreeNode*)malloc(sizeof(struct TreeNode));
node_l->val = atoi(token);
node->left = node_l;
queue[rear++] = node_l;
} else {
node->left = NULL;
}
token = strtok(NULL, ","); // right tree
if (token[0] != '#') {
struct TreeNode *node_r = (struct TreeNode*)malloc(sizeof(struct TreeNode));
node_r->val = atoi(token);
node->right = node_r;
queue[rear++] = node_r;
} else {
node->right = NULL;
}
}
return root;
}
void Inorder(struct TreeNode *root, int *ans, int *returnSize)
{
if (root != NULL) {
Inorder(root->left, ans, returnSize);
ans[(*returnSize)++] = root->val;
Inorder(root->right, ans, returnSize);
}
}
int *inorderTraversal(struct TreeNode *root, int *returnSize)
{
int *ans = (int *) malloc(sizeof(int) * 105);
*returnSize = 0;
Inorder(root, ans, returnSize);
return ans;
}
int main()
{
system("chcp 65001");
/* 用例输入的用一维数组表示的二叉树,输入用例记得在最后加四个# */
char nodeVal[] = {"3,9,20,#,#,15,7,#,#,#,#"};
int nodeNum = 5;
struct TreeNode* node = (TreeNode*)(malloc(sizeof(TreeNode)));
node = deserialize(nodeVal);
int* root = (int*) malloc(sizeof(int) * N);
int a;
root = inorderTraversal(node, &a);
for (int i = 0; i < nodeNum; i++) {
printf("%d ", root[i]);
}
return 0;
}
94-二叉树中序遍历-包含main的C程序
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转载自blog.csdn.net/zhangyuexiang123/article/details/125360608
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