PAT Advanced 1020 Tree Traversals (25 分)

1020 Tree Traversals (25 分)

 

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 (≤), 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

主要考查，中后遍历转前序，前中遍历建树，层序遍历树

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

struct TreeNode{
    int val;
    TreeNode *left;
    TreeNode *right;
};

vector<int> pre,in,post,ans;
queue<TreeNode*> que;

void preOrder(int root,int start,int end){
    if(start>end) return;
    int i=0;
    while(i<=end&&in[i]!=post[root]) i++;
    pre.push_back(post[root]);
    preOrder(root-1-end+i,start,i-1);
    preOrder(root-1,i+1,end);
}

TreeNode* buildTree(int root,int start,int end){
    if(start>end) return NULL;
    int i=0;
    TreeNode *t=new TreeNode();
    while(i<=end&&in[i]!=pre[root]) i++;
    t->val=pre[root];
    t->left=buildTree(root+1,start,i-1);
    t->right=buildTree(root+1+i-start,i+1,end);
    return t;
}

void levelOrder(TreeNode *tree){
    que.push(tree);
    while(!que.empty()){
        TreeNode *tmp=que.front();
        ans.push_back(tmp->val);
        que.pop();
        if(tmp->left!=NULL) que.push(tmp->left);
        if(tmp->right!=NULL) que.push(tmp->right);
    }

}
int main()
{
    int N;
    scanf("%d",&N);
    post.resize(N);in.resize(N);
    for(int i=0;i<N;i++) scanf("%d",&post[i]);
    for(int i=0;i<N;i++) scanf("%d",&in[i]);
    preOrder(N-1,0,N-1);
    TreeNode *tree=buildTree(0,0,N-1);
    levelOrder(tree);
    for(int i=0;i<ans.size();i++)
        if(i!=ans.size()-1) cout<<ans[i]<<" ";
        else cout<<ans[i];
    system("pause");
    return 0;
}

查看柳神博客，对代码修改，实际可以建立一个索引，这样就可以直接得到先序遍历的结果：

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
struct node{
    int index;
    int val;
};
bool cmp(node n1,node n2){
    return n1.index<n2.index;
}
vector<int> in,post;
vector<node> ans;

void preOrder(int root,int start,int end,int index){
    if(start>end) return;
    int i=0;
    while(i<=end&&in[i]!=post[root]) i++;
    ans.push_back({index,post[root]});
    preOrder(root-1-end+i,start,i-1,2*index+1);
    preOrder(root-1,i+1,end,2*index+2);
}
int main()
{
    int N;
    scanf("%d",&N);
    post.resize(N);in.resize(N);
    for(int i=0;i<N;i++) scanf("%d",&post[i]);
    for(int i=0;i<N;i++) scanf("%d",&in[i]);
    preOrder(N-1,0,N-1,0);
    sort(ans.begin(),ans.end(),cmp);
    for(int i=0;i<N;i++)
        if(i!=N-1) printf("%d ",ans[i].val);
        else printf("%d",ans[i].val);
    system("pause");
    return 0;
}