树与二叉树作业

1. 已知一个二叉树的中序遍历序列和后序遍历序列，求这棵树的前序遍历序列

【问题描述】
 已知一个二叉树的中序遍历序列和后序遍历序列，求这棵树的前序遍历序列。

【输入形式】
 一个树的中序遍历序列 该树后序遍历序列，中间用空格分开。输入序列中仅含有小写字母,且没有重复的字母

【输出形式】
 一个树的前序遍历序列

【样例输入】

dbeafcg debfgca

【样例输出】

abdecfg

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

struct TreeNode {
    char data;
    struct TreeNode* left;
    struct TreeNode* right;
};

int search(char in[], int start, int end, char value) {
    for (int i = start; i <= end; i++) {
        if (in[i] == value) {
            return i;
        }
    }
    return -1;
}

struct TreeNode* buildTree(char in[], char post[], int inStart, int inEnd, int* postIndex) {
    if (inStart > inEnd) {
        return NULL;
    }

    struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    node->data = post[(*postIndex)--];
    node->left = NULL;
    node->right = NULL;

    if (inStart == inEnd) {
        return node;
    }

    int rootIndex = search(in, inStart, inEnd, node->data);

    node->right = buildTree(in, post, rootIndex + 1, inEnd, postIndex);
    node->left = buildTree(in, post, inStart, rootIndex - 1, postIndex);

    return node;
}

void printPreorder(struct TreeNode* root) {
    if (root == NULL) {
        return;
    }
    printf("%c", root->data);
    printPreorder(root->left);
    printPreorder(root->right);
}

int main() {
    char inorder[100], postorder[100];
    scanf("%s %s", inorder, postorder);

    int len = strlen(inorder);
    int postIndex = len - 1;

    struct TreeNode* root = buildTree(inorder, postorder, 0, len - 1, &postIndex);

    printPreorder(root);
    printf("\n");

    return 0;
}

2. 非递归的中序遍历

【问题描述】

给定二叉树，返回它的中序遍历。

要求：使用栈实现，不可使用递归。

【输入形式】

一行，包含用空格分开的n个元素，每个元素为整数或者None（None表示空结点），依次表示按自顶向下层次遍历的二叉树结点。空结点对应的“孩子”（实际上不存在）不再用None表示。

【输出形式】

一行，二叉树的中序遍历，每个值间用空格隔开。

【样例输入】

1 2 3 4 5 6

【样例输出】

4 2 5 1 6 3

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

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

struct Stack {
    struct TreeNode **array;
    int top;
    int capacity;
};

struct Stack *createStack(int capacity) {
    struct Stack *stack = (struct Stack *)malloc(sizeof(struct Stack));
    stack->capacity = capacity;
    stack->top = -1;
    stack->array = (struct TreeNode **)malloc(stack->capacity * sizeof(struct TreeNode *));
    return stack;
}

int isEmpty(struct Stack *stack) {
    return stack->top == -1;
}

int isFull(struct Stack *stack) {
    return stack->top == stack->capacity - 1;
}

void push(struct Stack *stack, struct TreeNode *item) {
    if (isFull(stack)) return;
    stack->array[++stack->top] = item;
}

struct TreeNode *pop(struct Stack *stack) {
    if (isEmpty(stack)) return NULL;
    return stack->array[stack->top--];
}

void inorderTraversal(struct TreeNode *root) {
    if (root == NULL) return;

    struct Stack *stack = createStack(100);
    struct TreeNode *current = root;

    while (current != NULL || !isEmpty(stack)) {
        while (current != NULL) {
            push(stack, current);
            current = current->left;
        }
        current = pop(stack);
        printf("%d ", current->val);
        current = current->right;
    }

    free(stack->array);
    free(stack);
}

struct TreeNode *createTree(char input[][10], int n) {
    if (n == 0) return NULL;

    struct TreeNode **nodes = (struct TreeNode **)malloc(n * sizeof(struct TreeNode *));
    for (int i = 0; i < n; i++) {
        if (strcmp(input[i], "None") == 0) {
            nodes[i] = NULL;
        } else {
            nodes[i] = (struct TreeNode *)malloc(sizeof(struct TreeNode));
            nodes[i]->val = atoi(input[i]);
            nodes[i]->left = nodes[i]->right = NULL;
        }
    }

    for (int i = 0; i < n; i++) {
        if (nodes[i] != NULL) {
            int leftIndex = 2 * i + 1;
            int rightIndex = 2 * i + 2;
            if (leftIndex < n) nodes[i]->left = nodes[leftIndex];
            if (rightIndex < n) nodes[i]->right = nodes[rightIndex];
        }
    }

    struct TreeNode *root = nodes[0];
    free(nodes);
    return root;
}

void freeTree(struct TreeNode *root) {
    if (root == NULL) return;
    freeTree(root->left);
    freeTree(root->right);
    free(root);
}

int main() {
    char input[100][10];
    int n = 0;
    while (scanf("%s", input[n]) != EOF) {
        n++;
    }

    struct TreeNode *root = createTree(input, n);

    inorderTraversal(root);

    freeTree(root);

    return 0;
}

3. 判断完全二叉树

【问题描述】

给定一棵树的前序遍历和中序遍历，判断该树是否为完全二叉树
【输入形式】

一棵树的前序遍历和中序遍历
【输出形式】 

True or False

【样例输入】

2 4 8 10 6 11
8 4 10 2 11 6

【样例输出】

True

【样例说明】

【评分标准】

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

typedef struct Node {
    int data;
    struct Node *left, *right;
} Node;

Node* newNode(int data) {
    Node* node = (Node*)malloc(sizeof(Node));
    node->data = data;
    node->left = node->right = NULL;
    return node;
}

Node* buildTree(int pre[], int in[], int inStrt, int inEnd) {
    static int preIndex = 0;
    if (inStrt > inEnd) return NULL;
    Node* tNode = newNode(pre[preIndex++]);
    if (inStrt == inEnd) return tNode;
    int inIndex;
    for (inIndex = inStrt; inIndex <= inEnd; inIndex++) {
        if (in[inIndex] == tNode->data) break;
    }
    tNode->left = buildTree(pre, in, inStrt, inIndex - 1);
    tNode->right = buildTree(pre, in, inIndex + 1, inEnd);
    return tNode;
}

bool isCompleteBT(Node* root) {
    if (root == NULL) return true;

    bool end = false;
    Node* temp;
    Node* queue[1000]; // Changed to a regular array instead of dynamic allocation
    int front = 0, rear = 0;
    queue[rear++] = root;

    while (front < rear) {
        temp = queue[front++];

        if (temp->left) {
            if (end) return false;
            queue[rear++] = temp->left;
        } else {
            end = true;
        }

        if (temp->right) {
            if (end) return false;
            queue[rear++] = temp->right;
        } else {
            end = true;
        }
    }

    return true;
}

int main() {
    int preorder[100], inorder[100];
    int n = 0;
    char temp;
    while (scanf("%d", &preorder[n])) {
        temp = getchar();
        n++;
        if (temp == '\n') {
            break;
        }
    }
    for (int i = 0; i < n; i++) {
        scanf("%d", &inorder[i]);
    }

    Node* root = buildTree(preorder, inorder, 0, n - 1);
    if (isCompleteBT(root)) 
        printf("True\n");
    else 
        printf("False\n");
    return 0;
}

4. 二叉树遍历

【问题描述】

给定一棵N个节点的二叉树，输出其前序遍历，中序遍历，后序遍历，层次遍历。
【输入形式】

输入共N+1行。

第1行为一个整数N，描述节点个数。

其余N行按顺序描述第1，2，……，N个结点的左右子节点编号，0表示没有相应子节点。
【输出形式】

输出共4行，分别为前序遍历，中序遍历，后序遍历，层次遍历。
【样例输入】

10

8 0

4 1

0 0

6 9

0 0

3 7

0 0

0 0

5 10

0 0

【样例输出】

2 4 6 3 7 9 5 10 1 8 

3 6 7 4 5 9 10 2 8 1 

3 7 6 5 10 9 4 8 1 2 

2 4 1 6 9 8 3 7 5 10 

【数据范围】

保证输入的是合法的二叉树。

1<= N <= 10000.

#include <iostream>
#include <vector>
#include <queue>

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

TreeNode* build_tree(const std::vector<std::pair<int, int>>& nodes) {
    std::vector<TreeNode*> tree_nodes(nodes.size() + 1);
    std::vector<bool> is_root(nodes.size() + 1, true);
    for (int i = 1; i <= nodes.size(); ++i) {
        tree_nodes[i] = new TreeNode(i);
    }
    for (int i = 1; i <= nodes.size(); ++i) {
        tree_nodes[i]->left = nodes[i - 1].first ? tree_nodes[nodes[i - 1].first] : nullptr;
        tree_nodes[i]->right = nodes[i - 1].second ? tree_nodes[nodes[i - 1].second] : nullptr;
        if (nodes[i - 1].first) is_root[nodes[i - 1].first] = false;
        if (nodes[i - 1].second) is_root[nodes[i - 1].second] = false;
    }
    for (int i = 1; i <= nodes.size(); ++i) {
        if (is_root[i]) return tree_nodes[i];
    }
    return nullptr;
}


void preorder_traversal(TreeNode* root, std::vector<int>& result) {
    if (!root) return;
    result.push_back(root->val);
    preorder_traversal(root->left, result);
    preorder_traversal(root->right, result);
}

void inorder_traversal(TreeNode* root, std::vector<int>& result) {
    if (!root) return;
    inorder_traversal(root->left, result);
    result.push_back(root->val);
    inorder_traversal(root->right, result);
}

void postorder_traversal(TreeNode* root, std::vector<int>& result) {
    if (!root) return;
    postorder_traversal(root->left, result);
    postorder_traversal(root->right, result);
    result.push_back(root->val);
}

std::vector<int> level_order_traversal(TreeNode* root) {
    std::vector<int> result;
    std::queue<TreeNode*> q;
    if (root) q.push(root);
    while (!q.empty()) {
        TreeNode* node = q.front();
        q.pop();
        result.push_back(node->val);
        if (node->left) q.push(node->left);
        if (node->right) q.push(node->right);
    }
    return result;
}

int main() {
    int N;
    std::cin >> N;
    std::vector<std::pair<int, int>> nodes(N);
    for (int i = 0; i < N; ++i) {
        std::cin >> nodes[i].first >> nodes[i].second;
    }

    TreeNode* root = build_tree(nodes);

    std::vector<int> preorder, inorder, postorder, level_order;
    preorder_traversal(root, preorder);
    inorder_traversal(root, inorder);
    postorder_traversal(root, postorder);
    level_order = level_order_traversal(root);

    for (int val : preorder) std::cout << val << ' ';
    std::cout << '\n';
    for (int val : inorder) std::cout << val << ' ';
    std::cout << '\n';
    for (int val : postorder) std::cout << val << ' ';
    std::cout << '\n';
    for (int val : level_order) std::cout << val << ' ';
    std::cout << '\n';

    return 0;
}