剑指offer（18）——C++实现二叉树镜像

题目

操作给定的二叉树，将其变换为源二叉树的镜像

考察点

• 二叉树
• 二叉树遍历

解题思路

分析本质就是左右节点的交换。先前序遍历这棵树的每个结点，如果遍历到的结点有子结点，就交换它的两个子节点，当交换完所有的非叶子结点的左右子结点之后，就得到了树的镜像。

完整代码

/*18-二叉树的镜像*/
#include<iostream>
#include<stack>
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 *pRoot) {
		if (pRoot == NULL) return;
		
		TreeNode *temp;
		temp = pRoot->left;
		pRoot->left = pRoot->right;
		pRoot->right = temp;

		if (pRoot->left)
			Mirror(pRoot->left);
		if (pRoot->right)
			Mirror(pRoot->right);
	}
};
int main()
{
	TreeNode* A1 = new TreeNode(1); TreeNode* A2 = new TreeNode(2); TreeNode* A3 = new TreeNode(3);
	TreeNode* A4 = new TreeNode(4); TreeNode* A5 = new TreeNode(5); TreeNode* A6 = new TreeNode(6); TreeNode* A7 = new TreeNode(7);
	A1->left = A2; A1->right = A3; A2->left = A4; A2->right = A5; A3->left = A6; A3->right = A7;
	//打印原二叉树，前序遍历
	stack<TreeNode*> stack;
	stack.push(A1);
	while (!stack.empty())
	{
		A1 = stack.top();
		cout << A1->val << "";
		stack.pop();
		if (A1->right != NULL)
			stack.push(A1->right);
		if (A1->left != NULL)
			stack.push(A1->left);
	}
	cout << endl;
	Solution s;
	s.Mirror(A1);
	return 0;
}