1.题目要求按照后续和中序还原二叉树，然后进行层次遍历输出。
还原的函数为：
[c language=”++”]
class TreeNode{
public:
int val;
TreeNode* l, *r;
TreeNode(int x) :val(x), l(NULL), r(NULL){};
TreeNode() :val(0), l(NULL), r(NULL){};
};
TreeNode* build(vector<int>&postOrder, int postStart, int postEnd,
vector<int>&inOrder, int inStart, int inEnd, int&index, vector<TreeNode>&tree)
{
if (postStart == postEnd)
{//如果start和end相同，即这个是叶子节点
tree[index++] = TreeNode(postOrder[postEnd]);
return &tree[index – 1];
}
else if (postStart > postEnd) return NULL;
else
{
int rootVal = postOrder[postEnd];//后序遍历的末位是根
int rootPos = inStart;//根在中序遍历中的位置
for (; rootPos <= inEnd; rootPos++)
{
if (rootVal == inOrder[rootPos])
break;
}
int returnNode = index++;
tree[returnNode] = TreeNode(rootVal);
//注意newPostEnd = postStart + rootPos – inStart-1
tree[returnNode].l = build(postOrder, postStart, postStart + rootPos – inStart-1, inOrder, inStart, rootPos – 1, index, tree);
tree[returnNode].r = build(postOrder, postStart + rootPos – inStart , postEnd – 1, inOrder, rootPos + 1, inEnd, index, tree);
return &tree[returnNode];
}
}
[/c]
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
AC代码：
[c language=”++”]
//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
class TreeNode{
public:
int val;
TreeNode* l, *r;
TreeNode(int x) :val(x), l(NULL), r(NULL){};
TreeNode() :val(0), l(NULL), r(NULL){};
};
TreeNode* build(vector<int>&postOrder, int postStart, int postEnd,
vector<int>&inOrder, int inStart, int inEnd, int&index, vector<TreeNode>&tree)
{
if (postStart == postEnd)
{//如果start和end相同，即这个是叶子节点
tree[index++] = TreeNode(postOrder[postEnd]);
return &tree[index – 1];
}
else if (postStart > postEnd) return NULL;
else
{
int rootVal = postOrder[postEnd];//后序遍历的末位是根
int rootPos = inStart;//根在中序遍历中的位置
for (; rootPos <= inEnd; rootPos++)
{
if (rootVal == inOrder[rootPos])
break;
}
int returnNode = index++;
tree[returnNode] = TreeNode(rootVal);
//注意newPostEnd = postStart + rootPos – inStart-1
tree[returnNode].l = build(postOrder, postStart, postStart + rootPos – inStart-1, inOrder, inStart, rootPos – 1, index, tree);
tree[returnNode].r = build(postOrder, postStart + rootPos – inStart , postEnd – 1, inOrder, rootPos + 1, inEnd, index, tree);
return &tree[returnNode];
}
}
int main(void)
{
int nodeNum;
cin >> nodeNum;
vector<int> postOrder(nodeNum, 0);//后序
vector<int> inOrder(nodeNum, 0);//中序
for (int i = 0; i < nodeNum; i++)
{
cin >> postOrder[i];
}
for (int i = 0; i < nodeNum; i++)
{
cin >> inOrder[i];
}
vector<TreeNode> tree(nodeNum);//提前分配后内存空间，方便后面的连接
int index=0;
TreeNode*ans = build(postOrder, 0, nodeNum – 1, inOrder, 0, nodeNum – 1, index, tree);
int total = 0;
int total2 = 0;
queue<TreeNode*> q;
if (ans != NULL)
{
total = 1;
q.push(ans);
}
vector<int> outNum(0);
while (!q.empty())
{//进行BFS
while (total–)
{
TreeNode* front = q.front(); q.pop();
outNum.push_back(front->val);
if (front->l!=NULL)
{
q.push(front->l);
total2++;
}
if (front->r != NULL)
{
q.push(front->r);
total2++;
}
}
total = total2;
total2 = 0;
}
for (int i = 0; i < outNum.size(); i++)
{
cout << outNum[i];
if (i != outNum.size() – 1)
cout << " ";
}
return 0;
}
[/c]