PAT 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 7

Sample Output:

4 1 6 3 5 7 2

其实思路挺好想的，利用中序序列和后序序列就能构造出二叉树，然后对其进行层次遍历。只不过我不知道为什么第三个测试点是测什么的，过不了。。如果有同学做过或者看出来了，非常欢迎您帮我提出来，在这里先道声感谢。

#include <iostream>
#include <string>
#include <queue>
using namespace std;

struct TreeNode
{
int val;
TreeNode* l_child;
TreeNode* r_child;
TreeNode(int v = 0,TreeNode* l = NULL,TreeNode* r = NULL):val(v),l_child(l),r_child(r){}
};
typedef struct TreeNode* node;

node createTree(string post,string in)
{
node root = NULL;
if(post.size() == 0 || post.size() != in.size()) return root;
int index = in.find(post[post.size()-1]);
root = new struct TreeNode(post[post.size()-1] - '0',NULL,NULL);
int l_len = index;
int r_len = in.size() - 1 - index;
string l_post = post.substr(0,l_len);
string l_in = in.substr(0,l_len);
string r_post = post.substr(l_len,r_len);
string r_in = in.substr(index + 1,r_len);
root->l_child = createTree(l_post,l_in);
root->r_child = createTree(r_post,r_in);
return root;
}

void levelTraversal(node root)
{
queue<node> q;
if(root == NULL) return;
q.push(root);
bool tag = true;
node p = root;
while(!q.empty())
{
int value = q.front()->val;
if(tag)
tag = false;
else
printf(" ");
printf("%d",value);
if(p->l_child)
q.push(p->l_child);
if(p->r_child)
q.push(p->r_child);
q.pop();
p = q.front();
}
printf("\n");
}

int main()
{
int n;
cin>>n;
string post;
string in;
for(int i = 0; i < n; i++)
{
int v;
cin>>v;
post += v + '0';
}
for(int i = 0; i < n; i++)
{
int v;
cin>>v;
in += v + '0';
}
node root = NULL;
root = createTree(post,in);
levelTraversal(root);
return 0;
}