标签：遍历 queue seq bin name nta nod ace enc
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：由二叉树的中序遍历和后序遍历求出层序遍历的结果。
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<queue>
#include<stdlib.h>
using namespace std;
int N;
#define max 40
int postorder[max];
int inorder[max];
int cnt;
struct tree{
struct tree *left;
struct tree *right;
int num;
};
int searchValue(int num){
int i;
for(i=0;i<N;i++){
if(inorder[i]==num){
return i;
}
}
return -1;
}
tree * createTree(int left, int right){
if(left > right)return NULL;
int root = postorder[cnt];
cnt --;
int rootChild = searchValue(root);
tree* t= (tree*)malloc(sizeof(tree));
t->num = root;
if(left == right){
t->left = NULL;
t->right = NULL;
}
else{
t->right = createTree(rootChild+1,right);
t->left = createTree(left,rootChild-1);
//cout<<"hello world"<<endl;
}
return t;
}
void output(tree * t){
int levelorder[max];
queue<tree*>q_tree;
q_tree.push(t);
int i=0;
levelorder[i++]=t->num;
while(!q_tree.empty()){
tree* index = q_tree.front();
q_tree.pop();
if(index->left!=NULL){
q_tree.push(index->left);
levelorder[i++]=index->left->num;
}
if(index->right!=NULL){
q_tree.push(index->right);
levelorder[i++]=index->right->num;
}
}
int j;
for(j=0;j<i;j++){
if(j+1==i){
printf("%d\n",levelorder[j]);
}
else{
printf("%d ",levelorder[j]);
}
}
}
int main(){
scanf("%d",&N);
int i,j;
for(i=0;i<N;i++){
scanf("%d",&postorder[i]);
}
for(j=0;j<N;j++){
scanf("%d",&inorder[j]);
}
cnt = N-1;
tree* tt=createTree(0,N-1);
//cout<<"hello world"<<endl;
output(tt);
return 0;
}
标签：遍历 queue seq bin name nta nod ace enc
原文地址：http://www.cnblogs.com/grglym/p/7640533.html
