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  • 03-树3. Tree Traversals Again (25)将先序遍历和中序遍历转为后序遍历

    03-树3. Tree Traversals Again (25)

    题目来源:http://www.patest.cn/contests/mooc-ds/03-%E6%A0%913

    An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.


    Figure 1

    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

    题目实质是通过先序遍历和中序遍历建树,再后序遍历树。
    解题思路
    1. 通过输入建树
        Push操作代表新建一个节点,将其与父节点连接并同时压栈
        Pop操作,从栈顶弹出一个节点
    2. 后序遍历:递归实现
    代码如下:
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    
    #define STR_LEN 5
    #define MAX_SIZE 30
    
    typedef struct Node
    {
        int data;
        struct Node *left, *right;
    }* treeNode;
    
    treeNode Stack[MAX_SIZE];
    int values[MAX_SIZE];
    
    int num = 0;
    int top = -1;
    
    void Push(treeNode tn);
    treeNode Pop();
    treeNode Top();
    bool isEmpty();
    
    void PostOrderTraversal(treeNode root);
    
    int main()
    {
        int n;
        char operation[STR_LEN];
        treeNode father, root;
        bool findRoot = 0, Poped = 0;
    
        scanf("%d", &n);
        for (int i = 0; i < 2 * n; i++)
        {
            scanf("%s", operation);
            if (strcmp(operation, "Push") == 0)
            {
                int value;
                scanf("%d", &value);
                treeNode newNode;
                newNode = (treeNode)malloc(sizeof(struct Node));
                newNode->data = value;
                newNode->left = NULL;
                newNode->right = NULL;
                if (!findRoot)
                {
                    root = newNode;     //根节点
                    Push(newNode);
                    findRoot = 1;
                }
                else
                {
                    if (!Poped)     //如果前一个操作不是pop,则父节点为栈顶元素
                        father = Top();
                    if (father->left == NULL)
                        father->left = newNode;
                    else
                        father->right = newNode;
                    //printf("%d
    ", newNode->data);
                    Push(newNode);
                }
                Poped = 0;
            }
            else
            {
                father = Pop();
                Poped = 1;
            }
        }
        PostOrderTraversal(root);
    
        for (int i = 0; i < num-1; i++)
            printf("%d ", values[i]);
        printf("%d
    ", values[num-1]);
    
        return 0;
    }
    
    void PostOrderTraversal(treeNode root)
    {
        treeNode tn = root;
        if(tn)
        {
            PostOrderTraversal(tn->left);
            PostOrderTraversal(tn->right);
            values[num++] = tn->data;       //将后序遍历出的节点值存入数组便于格式化打印
        }
    }
    
    void Push(treeNode tn)
    {
        Stack[++top] = tn;
    }
    
    treeNode Pop()
    {
        return Stack[top--];
    }
    
    bool isEmpty()
    {
        return top == -1;
    }
    
    treeNode Top()
    {
        return Stack[top];
    }




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  • 原文地址:https://www.cnblogs.com/llhthinker/p/4748792.html
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