Algorithm Binary Tree

A question about how to build a binary tree from its front traval sequence and middle travel sequence

know everything about the binary tree can do it easily

Posted by Maizi on 2016-09-15 with time read

二叉树相关的算法问题

问题描述如下:

给定一棵二叉树的先序遍历和中序遍历的结果,怎么样获取其后续遍历的结果

算法描述:


算法分析:如果某个数组的元素顺序是某棵二叉树的先续遍历的顺序,那么从数组的第一个元素依次向后取元素可以重建这棵二叉树.

①从数组的第一个元素依次向后取元素重建这棵二叉树.

③后续遍历 ① 中重建的二叉树.

最后贴上代码:

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#include <stdio.h>
#include <stdlib.h>
typedef struct btree bt;
//二叉树的节点结构定义
struct btree {
    int _val;
    int _count;//此处优化结构,当元素重复时将该节点的计数加一
    bt *_lc;
    bt *_rc;
};
/**
 * 挂载节点
 * @param head 二叉树的根节点的指针
 * @param val 待插叙节点的值
 */
void insert(bt const * const head, int const val) {
    bt *node = malloc(sizeof(bt));
    node->_val = val;
    node->_count = 1;
    node->_lc = NULL;
    node->_rc = NULL;
    bt *tmp = head;
    while (1) {
        if (tmp->_val == val) {
            tmp->_count++;
            return;
        } else if (tmp->_val > val) {
            if (tmp->_lc == NULL)
                break;
            else
                tmp = tmp->_lc;
        } else {
            if (tmp->_rc == NULL)
                break;
            else
                tmp = tmp->_rc;
        }
    }
    if (tmp->_val > val)
        tmp->_lc = node;
    else
        tmp->_rc = node;
}
/**
 * 构建二叉树
 * @param arr 构建二叉树的输入-数组
 * @param len 数组的长度
 * @return 二叉树根节点的指针
 */
bt *buildTree(int const * const arr, size_t const len) {
    bt *head = malloc(sizeof(bt));
    head->_val = *(arr + len - 1);
    head->_count = 1;
    head->_lc = NULL;
    head->_rc = NULL;
    int i;
    for (i = 1; i < len; ++i) {
        insert(head, *(arr + i));
    }
    return head;
}
/**
 * 中序遍历(采用的递归算法)
 * @param node 当前的"根"节点指针
 */
void travelMidSeq(bt const * const node) {
    if (node->_lc != NULL)
        travelMidSeq(node->_lc);
    int i;
    int len = node->_count;
    for (i = 0; i < len; ++i)
        printf("%d ", node->_val);
    if (node->_rc != NULL)
        travelMidSeq(node->_rc);
}
/**
 * 释放资源(递归算法)---实质就是后续遍历的算法
 * @param node 当前需要处理的节点指针
 * @param arr 要排序的数组的指针
 */
void releaseTreeAndCheck(bt *node) {
    if (node->_lc != NULL)
        releaseTreeAndCheck(node->_lc);
    if (node->_rc != NULL)
        releaseTreeAndCheck(node->_rc);
    printf("%d ", node->_val);
    free(node);
}
/**
 * 打印数组
 * @param arr 数组的指针
 * @param len 数组的长度
 */
void print(int const * const arr, size_t const len) {
    int i;
    printf("the front travel sequence of the tree:\n");
    for (i = 0; i < len; ++i)
        printf("%d ", *(arr + i));
}
int main(void) {
    int arr[] = {8, 4, 2, 1, 3, 6, 5, 7, 12, 10, 9, 11, 14, 13, 15};
    size_t len = sizeof(arr) / sizeof(int);
    print(arr, len);
    bt *tree = buildTree(arr, len);
    printf("\nmiddle sequence travel result :\n");
    travelMidSeq(tree);
    printf("\nback sequence travel result :\n");
    releaseTreeAndCheck(tree);
    return 0;
}

附上测试结果:

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the front travel sequence of the tree:
8 4 2 1 3 6 5 7 12 10 9 11 14 13 15
middle sequence travel result :
1 2 3 4 5 6 7 9 10 11 12 13 14 15 15
back sequence travel result :
1 3 2 5 9 11 10 13 14 12 7 6 4 15


2016-09-15
— Maizi

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Algorithm Binary Tree
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