题解 | #二叉树遍历#

#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    char data;
    struct Node* leftNode;
    struct Node* rightNode;
} Node;

// 先序遍历拿到的是当前值,只需要左子树+1以及右子树+1
Node* create(Node* root, char str[], int* i) {
    char data = str[*i];
    if (data == '#' || data == '\0') {
        root = NULL;
        return root;
    } else {
        root = (Node*)malloc(sizeof(Node));
        root->data = data;
        (*i)++;
        root->leftNode = create(root->leftNode, str, i);
        (*i)++;
        root->rightNode = create(root->rightNode, str, i);
        return root;
    }
}

// 中序遍历
void LDR(Node* root) {
    if (root->leftNode != NULL) {
        LDR(root->leftNode);
    }
    printf("%c ", root->data);
    if (root->rightNode != NULL) {
        LDR(root->rightNode);
    }
}

int main() {
    Node* root;//定义一个根节点指针
    root = NULL;//将根节点指针赋值为空
    char str[100] = {0};
    int i = 0;
    scanf("%s", str);
    root = create(root, str, &i);//构建树
    LDR(root);
    return 0;
}