1020. Tree Traversals (25)
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 7Sample Output:
4 1 6 3 5 7 2
解题思路:根据二叉树的后序和中序序列来建树,并打印这颗二叉树的层序遍历。递归的去建树。后序最后一个节点是根节点,该根节点将中序序列分为左右子树两部分。递归去做。
代码如下:
#include<stdio.h> #include<string.h> #include<queue> #include<algorithm> using namespace std; struct Node { int data; Node* leftChild; Node* rightChild; }; int in[32], post[32]; vector<int> level; 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 countLeft = k - inL; root->leftChild = create(postL, postL+countLeft-1, inL, k-1); root->rightChild = create(postL+countLeft, postR-1, k+1, inR); return root; } void BFS(Node* root) { queue<Node*> que; que.push(root); while(!que.empty()) { Node *pre = que.front(); que.pop(); level.push_back(pre->data); if(pre->leftChild != NULL) que.push(pre->leftChild); if(pre->rightChild != NULL) que.push(pre->rightChild); } } int main() { freopen("D://input.txt", "r", stdin); int n; 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]); Node* root = create(0, n-1, 0, n-1); BFS(root); for(int i = 0; i < level.size(); i ++) { if(i == 0) printf("%d", level[i]); else printf(" %d", level[i]); } return 0; }