American Heritage
題目描述
Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear
tree in-order" and
tree pre-order" notations.
Your job is to create the `tree post-order" notation of a cow"s heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes.
Here is a graphical representation of the tree used in the sample input and output:
C / / B G / / A D H / E F The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree. The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree. The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root. ----------------------------------------------------------------------------------------------------------------------------
題目大意:
給出一棵二叉樹的前序遍歷 (preorder) 和中序遍歷 (inorder),求它的後序遍歷 (postorder)。
你需要知道的:
1:二叉樹的 相關定義可以在書上或者網上找到。
2:樣例 輸入輸出反映的二叉樹在上面。
輸入描述
Line 1:
The in-order representation of a tree.
Line 2:
The pre-order representation of that same tree.
輸出描述
A single line with the post-order representation of the tree.
樣例輸入
ABEDFCHG
CBADEFGH
樣例輸出
AEFDBHGC
#include<iostream> #include<string> using namespace std; char postorder[10001]; string inorder,preorder; int count=0; void dfs(int index,int left,int right)//index為根節點在preorder中的位置,left,right為樹的範圍 { if(left>right) { //postorder[count++]=preorder[index]; return ;} for(int i=0;i<inorder.size();i++) if(inorder[i]==preorder[index]) { dfs(index+1,left,i-1); dfs(index+i-left+1,i+1,right); postorder[count++]=inorder[i]; break; } } int main() { while(cin>>inorder>>preorder) { count=0; dfs(0,0,inorder.size()-1); for(int i=0;i<count;i++) cout<<postorder[i]; cout<<endl; } return 0;