合并二叉树、汉明距离、翻转二叉树

最新推荐文章于 2022-09-06 16:18:24 发布

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最新推荐文章于 2022-09-06 16:18:24 发布

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本文链接：https://blog.youkuaiyun.com/qq_41639336/article/details/93779027

本文详细介绍了LeetCode上的几个经典问题，包括如何合并两个二叉树，计算字符串之间的汉明距离，以及如何进行二叉树的翻转操作。内容涵盖解题思路和关键步骤。

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合并二叉树

//617合并二叉树
//Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
 
class Solution {
public:
	TreeNode* mergeTrees1(TreeNode* t1, TreeNode* t2) {
		//讨论为空情况
		if (t1 == NULL && t2 == NULL)
			return NULL;
		if (t1 == NULL)
			return t2;
		if (t2 == NULL)
			return t1;

		t1->val += t2->val;

		//t1左为空将t2的左拿过来，不为空遍历
		if (t1->left == NULL && t2->left != NULL)
		{
			t1->left = t2->left;
		}
		else
		{
			mergeTrees1(t1->left, t2->left);
		}

		//同上
		if (t1->right == NULL && t2->right != NULL)
		{
			t1->right = t2->right;
		}
		else
		{
			mergeTrees1(t1->right, t2->right);
		}

		return t1;
	}

	TreeNode* mergeTrees2(TreeNode* t1, TreeNode* t2) {
		if (t1 == NULL && t2 == NULL)
			return NULL;
		if (t1 == NULL)
			return t2;
		if (t2 == NULL)
			return t1;
		//if (t1 == NULL || t2 == NULL)
		//	return t1 == NULL ? t2 : t1;

		//此处直接获取返回值即可
		t1->left = mergeTrees2(t1->left, t2->left);
		t1->right = mergeTrees2(t1->right, t2->right);

		t1->val += t2->val;

		return t1;
	}
};

汉明距离

//461汉明距离
class Solution {
public:
	int hammingDistance(int x, int y) {
		int temp = x ^ y;
		int count = 0;

		while (temp)
		{
			if (temp & 1 == 1)
				++count;

			temp >>= 1;
		}

		return count;
	}

  //前一种方法必须完全查找temp的32位，本方法查找的次数取决于temp中1的个数，因此更高效
	int hammingDistance1(int x, int y) {
		int temp = x ^ y;
		int count = 0;

		while (temp)
		{
			++count;
			//temp-1后temp中最右边的1变为0，1右边的0变为1，按位与后其他位不变最后一个1变为0
			temp = temp & (temp - 1);
		}

		return count;
	}
};

翻转二叉树

// 226 翻转二叉树
//Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
	TreeNode* invertTree(TreeNode* root) {
		if (root == nullptr)
			return root;

		//先遍历到根节点，从下向上交换
		invertTree(root->left);
		invertTree(root->right);

		swap(root->left, root->right);

		return root;
	}

	//迭代写法
	TreeNode* invertTree1(TreeNode* root) {
		if (root == nullptr)
			return root;

		//用队列保存还没有交换过得结点
		queue<TreeNode*> tnq;
		tnq.push(root);

		while (!tnq.empty())
		{
			//拿出一个需要交换的结点
			TreeNode* now = tnq.front();
			tnq.pop();
			swap(now->left, now->right);

			//将当前结点不为空的子树存入队列中
			if (now->left != nullptr)
				tnq.push(now->left);
			if (now->right != nullptr)
				tnq.push(now->right);
		}

		return root;
	}
};

题目前序号为LeetCode上的题号，部分解法思路来自LeetCode