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：已知一棵树的后序遍历（postorder）和中序遍历（inoder），求这棵树按照层输出的结果（bfs输出）
题目思路：
首先先说明这道题我是网上搜了题解的，代码参考了
http://blog.csdn.net/xyt8023y/article/details/46273967
这篇博客，文章里思路介绍的很清楚，大家可以转到这片博客。下面站在我的角度说一说这道题。
首先这是一道树的题，而且是二叉树，二叉树的题最核心最核心的思想我觉得就是递归，所以不难想到这道题一定会用到递归。
怎么用呢？我们知道，针对每一个子树来说后序遍历的最末尾就是该树的根，再通过这个数字在中序遍历中的位置，可以将这棵子树再划分为左右两个子树。
那么程序怎么实现呢？首先递归一定要有参数，所以就以对于中序遍历来说左右两个子树的端点来做参数，同时需要一个current变量全局变量来作为当前子树的根的后序遍历索引。
当构建出来树之后，就是对树bfs遍历即可（建议背住）：
#include <iostream> #include<vector> #include<queue> using namespace std; vector<int>postorder; vector<int>inorder; int current; struct tree { tree* left; tree* right; int root; }; int get_index(int c) { for(int i=0;i<(int)inorder.size();i++) { if(c==inorder[i]) return i; } return -1; } tree* build(int left,int right) { if(left>right)//递归终止条件 return NULL; tree* node=(tree*)malloc(sizeof(struct tree));//给节点分配空间 int curRoot=postorder[current]; current--; int index=get_index(curRoot);//获取当前节点在中序遍历的索引 node->root=curRoot; if(right==left)//叶子节点 { node->right=NULL; node->left=NULL; } else { node->right=build(index+1,right); node->left=build(left,index-1);//注意顺序 } return node; } void bfs(tree *t) { bool first=true; queue<tree*>q; q.push(t); while(!q.empty()) { tree* temp=q.front(); q.pop(); if(first) { printf("%d",temp->root); first= false; } else printf(" %d",temp->root); if(temp->left!=NULL) { q.push(temp->left); } if(temp->right!=NULL) { q.push(temp->right); } } } int main() { int n; postorder.resize(n); inorder.resize(n); scanf("%d",&n); current=n-1; for(int i=0;i<n;i++) { int temp; scanf("%d",&temp); postorder.push_back(temp); } for(int i=0;i<n;i++) { int temp; scanf("%d",&temp); inorder.push_back(temp); } tree* t=build(0,n-1); bfs(t); return 0; }
