【PAT Advanced Level】1020. Tree Traversals (25)

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这题给定后序和中序遍历的结点顺序，要求层序遍历的结点顺序。

可以通过后序和中序构造出这棵树，然后再BFS给出层序遍历。构造的规则为：

1. 找到后序遍历的最后一个元素，这个元素肯定是整个树的根。

2. 根据上面找到的元素，将中序遍历的序列分为两边，左边肯定是根的左子树，右边是根的右子树

3. 递归左子树和右子树

#include <iostream>#include <vector>#include <algorithm>#include <fstream>#include <queue>using namespace std;int k;vector<int> post;vector<int> in;vector<int> reslut;struct tree_node{int data;tree_node *left;tree_node *right;tree_node() {left = NULL;right = NULL;}};tree_node *root; tree_node *preOrder(vector<int>::iterator pBegin, vector<int>::iterator pEnd, vector<int>::iterator iBegin, vector<int>::iterator iEnd, tree_node *root){if(pBegin == pEnd || iBegin == iEnd)return NULL;int tmp = *(pEnd - 1);root = new tree_node;root->data = tmp;vector<int>::iterator index = find(iBegin, iEnd, tmp);root->left = preOrder(pBegin, pBegin + (index - iBegin), iBegin, index, root->left);root->right = preOrder(pBegin + (index - iBegin), pEnd - 1, index + 1, iEnd, root->right);return root;} void BSF() { queue<tree_node *> q; q.push(root); while (!q.empty()) { tree_node *tmp = q.front(); q.pop(); reslut.push_back(tmp->data); if(tmp->left != NULL) q.push(tmp->left); if(tmp->right != NULL) q.push(tmp->right); } }int main(){//fstream cin("a.txt");cin>>k;int tmp = k;post.push_back(0);in.push_back(0);while (tmp--){int node;cin>>node;post.push_back(node);}tmp = k;while (tmp--){int node;cin>>node;in.push_back(node);}root = preOrder(post.begin() + 1, post.end(), in.begin() + 1, in.end(), root);BSF();for(int i = 0; i < reslut.size(); i++){if(i == 0)cout<<reslut[i];elsecout<<" "<<reslut[i];}cout<<endl;}