寻找最低公共父节点

最新推荐文章于 2021-05-14 10:12:34 发布

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本文讨论了最低公共父节点的概念，指出其需要满足的条件，并提出了思路。在实现过程中，通过设置标志位来判断节点在左子树或右子树。文章提及了特殊情况的处理，如节点相等、节点可能不是叶子节点，以及节点是否一定存在于树中。同时，指出了原始代码的效率问题，提出使用单个递归过程来优化解决方案。

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最低公共父节点，意思很好理解。

思路1：最低公共父节点满足这样的条件：两个节点分别位于其左子树和右子树，那么定义两个bool变量，leftFlag和rightFlag，如果在左子树中，leftFlag为true，如果在右子树中，rightFlag为true，仅当leftFlag == rightFlag == true时，才能满足条件

#include <iostream>

using namespace std;

struct Node
{
	Node(int i = 0, Node *pLeft = NULL, Node *pRight = NULL) : data(i), left(pLeft),
		right(pRight) {}
	Node *left;
	Node *right;
	int data;
};

Node *constructNode(Node **pNode1, Node **pNode2)
{
	Node *node12 = new Node(12);
	Node *node11 = new Node(11);
	Node *node10 = new Node(10);
	Node *node9 = new Node(9, NULL, node12);
	Node *node8 = new Node(8, node11, NULL);
	Node *node7 = new Node(7);
	Node *node6 = new Node(6);
	Node *node5 = new Node(5, node8, node9);
	Node *node4 = new Node(4, node10);
	Node *node3 = new Node(3, node6, node7);
	Node *node2 = new Node(2, node4, node5);
	Node *node1 = new Node(1, node2, node3);

	*pNode1 = node6;
	*pNode2 = node12;

	return node1;
}

bool isNodeIn(Node *root, Node *node1, Node *node2)
{
	if (node1 == NULL || node2 == NULL)
	{
		throw("invalid node1 and node2");
		return false;
	}
	if (root == NULL)
		return false;

	if (root == node1 || root == node2)
	{
		return true;
	}
	else
	{
		return isNodeIn(root->left, node1, node2) || isNodeIn(root->right, node1, node2);
	}
}

Node *lowestFarther(Node *root, Node *node1, Node *node2)
{
	if (root == NULL || node1 == NULL || node2 == NULL || node1 == node2)
	{
		return NULL;
	}
	
	bool leftFlag = false;
	bool rightFlag = false;
	leftFlag = isNodeIn(root->left, node1, node2);
	rightFlag = isNodeIn(root->right, node1, node2);

	if (leftFlag == true && rightFlag == true)
	{
		return root;
	}
	else if (leftFlag == true)
	{
		return lowestFarther(root->left, node1, node2);
	}
	else
	{
		return lowestFarther(root->right, node1, node2);
	}
}

void main()
{
	Node *node1 = NULL;
	Node *node2 = NULL;
	Node *root = constructNode(&node1, &node2);

	cout << "node1: " << node1->data << endl;
	cout << "node2: " << node2->data << endl;
	cout << "root: " << root->data << endl;

	Node *father = lowestFarther(root, node1, node2);

	if (father == NULL)
	{
		cout << "no common father" << endl;
	}
	else
	{
		cout << "father: " << father->data << endl;
	}
}

思考：在面试的时候，得和面试官沟通，考虑以下情形

1. node1和node2指向同一节点，这个如何处理

2. node1或node2有不为叶子节点的可能性吗

3. node1或node2一定在树中吗

还要考虑一个效率问题，上述代码中用了两个递归函数，而且存在不必要的递归过程，仔细思考，其实一个递归过程足以解决此问题

#include <iostream>

using namespace std;

struct Node
{
	Node(int i = 0, Node *pLeft = NULL, Node *pRight = NULL) : data(i),
		left(pLeft), right(pRight) {}
	int data;
	Node *left;
	Node *right;
};

Node *constructNode(Node **pNode1, Node **pNode2)  
{  
	Node *node12 = new Node(12);  
	Node *node11 = new Node(11);  
	Node *node10 = new Node(10);  
	Node *node9 = new Node(9, NULL, node12);  
	Node *node8 = new Node(8, node11, NULL);  
	Node *node7 = new Node(7);  
	Node *node6 = new Node(6);  
	Node *node5 = new Node(5, node8, node9);  
	Node *node4 = new Node(4, node10);  
	Node *node3 = new Node(3, node6, node7);  
	Node *node2 = new Node(2, node4, node5);  
	Node *node1 = new Node(1, node2, node3);  

	*pNode1 = node6;  
	*pNode2 = node5;  

	return node1;  
}

bool lowestFather(Node *root, Node *node1, Node *node2, Node *&dest)
{
	if (root == NULL || node1 == NULL || node2 == NULL || node1 == node2)
		return false;
	if (root == node1 || root == node2)
		return true;

	bool leftFlag = lowestFather(root->left, node1, node2, dest);
	bool rightFlag = lowestFather(root->right, node1, node2, dest);
	
	if (leftFlag == true && rightFlag == true)
	{
		dest = root;
	}
	if (leftFlag == true || rightFlag == true)
		return true;
}

int main()
{
	Node *node1 = NULL;
	Node *node2 = NULL;
	Node *root = constructNode(&node1, &node2);

	bool flag1 = false;
	bool flag2 = false;
	Node *dest = NULL;
	bool flag = lowestFather(root, node1, node2, dest);

	if (dest != NULL)
	{
		cout << "lowest common father: " << dest->data << endl;
	}
	else
	{
		cout << "no common father!" << endl;
	}

	return 0;
}

换一种方式的写法：

#include <iostream>

using namespace std;

struct Node
{
	Node(int i = 0, Node *pLeft = NULL, Node *pRight = NULL) : data(i),
		left(pLeft), right(pRight) {}
	int data;
	Node *left;
	Node *right;
};

Node *constructNode(Node **pNode1, Node **pNode2)  
{  
	Node *node12 = new Node(12);  
	Node *node11 = new Node(11);  
	Node *node10 = new Node(10);  
	Node *node9 = new Node(9, NULL, node12);  
	Node *node8 = new Node(8, node11, NULL);  
	Node *node7 = new Node(7);  
	Node *node6 = new Node(6);  
	Node *node5 = new Node(5, node8, node9);  
	Node *node4 = new Node(4, node10);  
	Node *node3 = new Node(3, node6, node7);  
	Node *node2 = new Node(2, node4, node5);  
	Node *node1 = new Node(1, node2, node3);  

	*pNode1 = node11;  
	*pNode2 = node12;  

	return node1;  
}

Node* lowestFather(Node *root, Node *node1, Node *node2)
{
	if (root == NULL || node1 == NULL || node2 == NULL || node1 == node2)
		return NULL;
	if (root == node1 || root == node2)
		return root;

	Node* leftFlag = lowestFather(root->left, node1, node2);
	Node* rightFlag = lowestFather(root->right, node1, node2);

	if (leftFlag == NULL)
		return rightFlag;
	else if (rightFlag == NULL)
		return leftFlag;
	else
		return root;
}

int main()
{
	Node *node1 = NULL;
	Node *node2 = NULL;
	Node *root = constructNode(&node1, &node2);

	bool flag1 = false;
	bool flag2 = false;
	Node *dest = NULL;
	Node* flag = lowestFather(root, node1, node2);

	if (flag != NULL)
	{
		cout << "lowest common father: " << flag->data << endl;
	}
	else
	{
		cout << "no common father!" << endl;
	}

	return 0;
}