• PAT006 Tree Traversals Again


    题目:

    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

    分析: 主要是根据输入创建一个二叉树,然后进行后续遍历

    代码:

    #pragma mark -Tree Traversals Again
    #include <stdio.h>
    
    typedef struct traversalTreeNode {
        int value;
        struct traversalTreeNode *left;
        struct traversalTreeNode *right;
    } TraversalTreeNode;
    
    int flag;
    
    TraversalTreeNode *createTraversalTreeNode(int value)
    {
        TraversalTreeNode *node = (TraversalTreeNode *)malloc(sizeof(TraversalTreeNode));
        node->value = value;
        node->left = NULL;
        node->right = NULL;
        return node;
    }
    
    void postorderTraversal(TraversalTreeNode *head)
    {
        if (head) {
            postorderTraversal(head->left);
            postorderTraversal(head->right);
            
            if (flag == 0) {
                printf("%d", head->value);
                flag = 1;
            } else {
                printf(" %d", head->value);
            }
        }
    }
    
    int main()
    {
        int nodeNum = 0;
        scanf("%d", &nodeNum);
        
        int operationCount = 2 * nodeNum;
        TraversalTreeNode *a[30];
        
        int top = -1;
        // 第一个节点肯定是PUSH
        int index = -1;
        scanf("%*s %d", &index);
        TraversalTreeNode *head = createTraversalTreeNode(index);
        a[0] = head;
        top = 0;
        
        TraversalTreeNode *popItem = NULL;
        for (int i = 1; i < operationCount; i++) {
            int index = -1;
            char str[4];
            scanf("%s", str);
            unsigned long len = strlen(str);
            if (len >= 4) {
                scanf("%d", &index);
                TraversalTreeNode *newNode = createTraversalTreeNode(index);
                if (popItem) {
                    if (!popItem->left) {
                        popItem->left = newNode;
                    } else {
                        popItem->right = newNode;
                    }
                } else {
                    if (!a[top]->left) {
                        a[top]->left = newNode;
                    } else {
                        a[top]->right = newNode;
                    }
                }
                
                top++;
                a[top] = newNode;
                popItem = NULL;
            } else {
                popItem = a[top];
                a[top] = NULL;
                top--;
            }
        }
        
        flag = 0;
        postorderTraversal(head);
    }

    运行结果:

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  • 原文地址:https://www.cnblogs.com/liufeng24/p/4419588.html
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