二叉树--3-二叉树镜像

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:

剑指Offer(十八):二叉树的镜像

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:

剑指Offer(十八):二叉树的镜像

#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;
}

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转载自blog.csdn.net/ax_hacker/article/details/81026657