PAT 1020 Tree Traversals根据中序，后序遍历结果重新建树

最新推荐文章于 2022-02-26 19:51:19 发布

adventural

最新推荐文章于 2022-02-26 19:51:19 发布

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分类专栏： PAT 文章标签：二叉树的遍历

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本文探讨了如何从已知的二叉树中序和后序遍历序列中，通过递归方法构建二叉树，并实现层序遍历输出。通过分析后序遍历最后一个元素确定根节点，进而划分中序序列，构建左右子树，最终完成二叉树的层序遍历。

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1020 Tree Traversals

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification

Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2

题目大意：已知二叉树的中序和后序遍历结果，求层序遍历结果
思路：后序的最后一位即是树的根节点，根节点把中序序列化为两部分，左边是左子树，右边是右子树，递归建树

AC代码

#include <cstdio>
#include <queue>

using namespace std;

struct treenode{
    int num;
    struct treenode* left;
    struct treenode* right;
};

struct treenode* Creat(int Pleft, int Pright, int Ileft, int Iright);

int N, post[30], in[30];
queue<treenode*> q;

int main()
{
    scanf("%d", &N);
    for (int i=0; i<N; i++)
        scanf("%d", &post[i]);
    for (int i=0; i<N; i++)
        scanf("%d", &in[i]);
    treenode* T;
    T = Creat(0,N-1,0,N-1);
    q.push(T);
    bool tmp = true;
    while(!q.empty())
    {
        if (tmp==true)
            printf("%d",q.front()->num);
        else
            printf(" %d",q.front()->num);
        if (q.front()->left != NULL)
            q.push(q.front()->left);
        if (q.front()->right != NULL)
            q.push(q.front()->right);
        q.pop();
        tmp = false;
    }
    return 0;
}

struct treenode* Creat(int Pleft, int Pright, int Ileft, int Iright)
{
    if (Pleft > Pright)
        return NULL;
    treenode* T;
    int index, leftlen, rightlen;
    T = new treenode;
    T->num = post[Pright];
    for (index=Ileft; index<=Iright; index++)
    {
        if (in[index] == post[Pright])
            break;
    }
    leftlen = index-Ileft;
    T->left = Creat(Pleft, Pleft+leftlen-1, Ileft, index-1);
    rightlen = Iright-index;
    T->right = Creat(Pright-rightlen, Pright-1, index+1, Iright);
    return T;
}