1.和leetcode中的Invert Binary Tree（easy）题目一样。
2.利用递归函数，每次都把左右子树翻转即可。
翻转函数如下：
[c language=”++”]
void invertTree(TreeNode* root)
{
if (root != NULL)
{
swap(root->l, root->r);
invertTree(root->l);
invertTree(root->r);
}
}
[/c]
The following is from Max Howell @twitter:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
Now it’s your turn to prove that YOU CAN invert a binary tree!
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree — and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the left and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.
Sample Input:
8 1 - - - 0 - 2 7 - - - - 5 - 4 6
Sample Output:
3 7 2 6 4 0 5 1 6 5 7 4 3 2 0 1
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;
struct TreeNode{
int val;
TreeNode*l, *r;
TreeNode() :val(-1),l(NULL), r(NULL){};
};
void invertTree(TreeNode* root)
{
if (root != NULL)
{
swap(root->l, root->r);
invertTree(root->l);
invertTree(root->r);
}
}
void levelOrder(TreeNode*root,vector<int>&ans)
{
queue<TreeNode*> q;
if (root != NULL)
{
q.push(root);
int count1 = 1;
int count2 = 0;
while (!q.empty())
{
for (int i = 0; i < count1; i++)
{
TreeNode* head = q.front(); q.pop();
ans.push_back(head->val);
if (head->l != NULL)
{
q.push(head->l);
count2++;
}
if (head->r != NULL)
{
q.push(head->r);
count2++;
}
}
count1 = count2;
count2 = 0;
}
}
}
void inOrder(TreeNode*root, vector<int>&ans)
{
if (root != NULL)
{
inOrder(root->l, ans);
ans.push_back(root->val);
inOrder(root->r, ans);
}
}
int main(void)
{
int sum;
cin >> sum;
vector<TreeNode> tree(sum);
vector<int> degree(sum,0);
for (int i = 0; i < sum; i++)
{
tree[i].val = i;
char a, b;
cin >> a >> b;
if (a!=’-‘)
{
tree[i].l = &tree[a – ‘0’];
degree[a – ‘0’]++;
}
if (b != ‘-‘)
{
tree[i].r = &tree[b – ‘0’];
degree[b – ‘0’]++;
}
}
TreeNode* root=NULL;
for (int i = 0; i < sum; i++)
{
if (degree[i] == 0)
{
root = &tree[i];
break;
}
}
invertTree(root);
vector<int> ans1(0);
vector<int> ans2(0);
levelOrder(root, ans1);
inOrder(root, ans2);
for (int i = 0; i < ans1.size(); i++)
{
cout << ans1[i];
if (i != ans1.size() – 1)
cout << " ";
}
cout << endl;
for (int i = 0; i < ans2.size(); i++)
{
cout << ans2[i];
if (i != ans2.size() – 1)
cout << " ";
}
cout << endl;
return 0;
}
[/c]