PAT A1020 Tree Traversals （25 分）——建树，层序遍历-白红宇

PAT A1020 Tree Traversals （25 分）——建树，层序遍历

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发布时间：2019-06-14

本文共 2154 字，大约阅读时间需要 7 分钟。

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

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:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               using namespace std;const int maxn = 10000;int n;int post[maxn], in[maxn];struct node { int data; node* left; node* right;};void layerorder(node* root) { queue
                
                  q; q.push(root); int count = 0; while (!q.empty()) { node* now = q.front(); q.pop(); printf("%d", now->data); count++; if (now->left != NULL) q.push(now->left); if (now->right != NULL) q.push(now->right); if (count != n) printf(" "); }}node* create(int postl, int postr, int inl, int inr) { if (postl > postr) { return NULL; } node* root = new node; root->data = post[postr]; int k; for (k = inl; k <= inr; k++) { if (in[k] == post[postr]) { break; } } int leftnum = k - inl; root->left = create(postl, postl + leftnum - 1, inl, k-1); root->right = create(postl + leftnum, postr - 1, k + 1, inr); return root;}int main() { cin >> n; for (int i = 0; i < n; i++) { cin>>post[i]; } for (int i = 0; i < n; i++) { cin >> in[i]; } node* root = create(0, n - 1, 0, n - 1); layerorder(root); system("pause");}

注意点：考察基本的二叉树遍历，难点在递归上，想清楚了递归边界和递归式就简单了。二叉树的遍历及建树还需要巩固。

转载于:https://www.cnblogs.com/tccbj/p/10392622.html

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