03-树3 Tree Traversals Again (25分)

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

解题思路

通过观察Push和Pop的顺序可以发现，Push的顺序和先序遍历一样，Pop的顺序和中序遍历一样。因此可以先通过两个数组分别存储先序遍历序列和中序遍历序列，再通过这两个数组构建二叉树，最后再后序遍历即可。

代码

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

#define MAXSIZE 30

struct StackNode {
    int data;
    struct StackNode *next;
};
typedef struct StackNode *Stack;

Stack createStack() {
    Stack stack = (Stack) malloc(sizeof(struct StackNode));
    stack->next = NULL;
    return stack;
}

void push(Stack stack, int data) {
    Stack node = createStack();
    node->data = data;
    node->next = stack->next;
    stack->next = node;
}

int pop(Stack stack) {
    if (!stack->next) return -1;
    Stack top = stack->next;
    int data = top->data;
    stack->next = top->next;
    free(top);
    return data;
}

struct TreeNode {
    int data;
    struct TreeNode *left;
    struct TreeNode *right;
};
typedef struct TreeNode *Tree;

int n, num = 0;
int preOrder[MAXSIZE], inOrder[MAXSIZE];

Tree createTree(int preL, int preR, int inL, int inR) {
    if (preL > preR) return NULL;
    Tree root = (Tree) malloc(sizeof(struct TreeNode));
    root->data = preOrder[preL];
    int rootIndex = inL;
    for ( ; rootIndex <= inR; rootIndex++) {
        if (preOrder[preL] == inOrder[rootIndex]) break;
    }
    int leftNums = rootIndex - inL;
    root->left = createTree(preL + 1, preL + leftNums, inL, rootIndex - 1);
    root->right = createTree(preL + leftNums + 1, preR, rootIndex + 1, inR);
    return root;
}

void postOrder(Tree root) {
    if (root == NULL) return;
    postOrder(root->left);
    postOrder(root->right);
    printf("%d", root->data);
    num++;
    if (num < n) printf(" ");
}

int main() {
    scanf("%d\n", &n);
    Stack stack = createStack();
    char input[5];
    int data, preIndex = 0, inIndex = 0;
    for (int i = 0; i < 2 * n; i++) {
        scanf("%s", input);
        if (strcmp(input, "Push") == 0) {
            scanf("%d\n", &data);
            push(stack, data);
            preOrder[preIndex++] = data;
        } else {
            inOrder[inIndex++] = pop(stack);
        }
    }
    Tree root = createTree(0, n - 1, 0, n - 1);
    postOrder(root);
    return 0;
}

03-树3 Tree Traversals Again (25分)

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原文地址：https://www.cnblogs.com/AndyHY-Notes/p/12540311.html