转载:http://cuijiahua.com/blog/2017/12/basis_18.html
题目
操作给定的二叉树,将其变换为源二叉树的镜像。
如下图所示:
思路
先交换根节点的两个子结点之后,我们注意到值为10、6的结点的子结点仍然保持不变,因此我们还需要交换这两个结点的左右子结点。做完这两次交换之后,我们已经遍历完所有的非叶结点。此时变换之后的树刚好就是原始树的镜像。交换示意图如下所示:
#include <iostream>
using namespace std;
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
class Solution {
public:
void mirror(TreeNode *root)
{
if (root == nullptr || (root->left == nullptr && root->right == nullptr))
return;
if (root->left)
mirror(root->left);
if (root->right)
mirror(root->right);
TreeNode *temp = root->left;
root->left = root->right;
root->right = temp;
}
};
int main()
{
Solution s;
TreeNode *head = new TreeNode(8);
TreeNode *head1 = new TreeNode(6);
head->left = head1;
TreeNode *head2 = new TreeNode(10);
head->right = head2;
TreeNode *head3 = new TreeNode(5);
head2->left = head3;
s.mirror(head);
cout << head->val << head->left->val << head->right->val << head->left->right->val << endl;
system("pause");
return 0;
}