C++算法之 求二叉树两个节点的最低公共节点

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方法1：递归方法：

（1）如果两个节点分别在根节点的左子树和右子树，则返回根节点
（2）如果两个节点都在左子树，则递归处理左子树；如果两个节点都在右子树，则递归处理右子树

bool FindNode(BTree* pRoot, BTree* pNode)
{
	if (pRoot == NULL || pNode == NULL)
	{
		return false;
	}
	if (pRoot == pNode)
	{
		return true;
	}

	bool found = FindNode(pRoot->m_nLeft, pNode);
	if (!found)
	{
		found = FindNode(pRoot->m_nRight,pNode);
	}

	return found;
}

BTree* GetCommentParent(BTree* pRoot, BTree* pNode1, BTree* pNode2)
{
	if (FindNode(pRoot->m_nLeft,pNode1))
	{
		if (FindNode(pRoot->m_nRight, pNode2))
		{
			return pRoot;//如果两个节点分别在根节点的左子树和右子树，则返回根节点
		}
		else//如果两个节点都在左子树，则递归处理左子树
		{
			return GetCommentParent(pRoot->m_nLeft,pNode1,pNode2);
		}
	}
	else
	{
		if (FindNode(pRoot->m_nLeft,pNode2))
		{
			return pRoot;//如果两个节点分别在根节点的左子树和右子树，则返回根节点
		}
		else//如果两个节点都在右子树，则递归处理右子树
		{
			return GetCommentParent(pRoot->m_nRight,pNode1,pNode2);
		}
	}
}

方法2：

非递归解法：
先求从根节点到两个节点的路径，然后再比较对应路径的节点就行，最后一个相同的节点也就是他们在二叉树中的最低公共祖先节点

bool GetNodePath(BTree* pRoot, BTree* pNode, list<BTree*> & path)
{
	if (pRoot == pNode)
	{
		path.push_back(pRoot);
		return true;
	}
	if (pRoot == NULL)
	{
		return false; 
	}

	path.push_back(pRoot);
	bool found = false;
	found = GetNodePath(pRoot->m_nLeft,pNode,path);
	if (!found)
	{
		found = GetNodePath(pRoot->m_nRight,pNode,path);
	}
	if (!found)
	{
		path.pop_back();
	}
	return found;
}

BTree* GetCommentParent2(BTree* pRoot,BTree* pNode1,BTree* pNode2)
{
	if (pRoot == NULL || pNode1 == NULL || pNode2 == NULL)
	{
		return NULL;
	}
	list<BTree*> path1;
	list<BTree*> path2;
	bool bResult1 = GetNodePath(pRoot,pNode1,path1);
	bool bResult2 = GetNodePath(pRoot,pNode2,path2);
	if (!bResult1 || !bResult2)
	{
		return NULL;
	}

	list<BTree*>::const_iterator iter1 = path1.begin();
	list<BTree*>::const_iterator iter2 = path2.begin();
	BTree* pCommentParent = NULL;
	while (iter1 != path1.end() && iter2 != path2.end())
	{
		if (*iter1 == *iter2)
		{
			pCommentParent = *iter1;
		}
		
		iter1++;
		iter2++;
	}

	return pCommentParent;
}

完整测试代码：

// FindCommentParent.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <list>
#include <queue>
using namespace std;

//节点的数据结构
class BTree
{
public:
	int       m_nValue;
	BTree*    m_nLeft;
	BTree*    m_nRight;
public:
	BTree(int value)
	{
		m_nValue = value;
	}
};

//二叉树的插入实现
void Insert(int value, BTree* &root)
{
	if (root == NULL)
	{
		root = new BTree(value);
	}
	else if(value < root->m_nValue)
		Insert(value,root->m_nLeft);
	else if(value > root->m_nValue)
		Insert(value,root->m_nRight);
	else
		;
}

bool FindNode(BTree* pRoot, BTree* pNode)
{
	if (pRoot == NULL || pNode == NULL)
	{
		return false;
	}
	if (pRoot == pNode)
	{
		return true;
	}

	bool found = FindNode(pRoot->m_nLeft, pNode);
	if (!found)
	{
		found = FindNode(pRoot->m_nRight,pNode);
	}

	return found;
}

BTree* GetCommentParent(BTree* pRoot, BTree* pNode1, BTree* pNode2)
{
	if (FindNode(pRoot->m_nLeft,pNode1))
	{
		if (FindNode(pRoot->m_nRight, pNode2))
		{
			return pRoot;//如果两个节点分别在根节点的左子树和右子树，则返回根节点
		}
		else//如果两个节点都在左子树，则递归处理左子树
		{
			return GetCommentParent(pRoot->m_nLeft,pNode1,pNode2);
		}
	}
	else
	{
		if (FindNode(pRoot->m_nLeft,pNode2))
		{
			return pRoot;//如果两个节点分别在根节点的左子树和右子树，则返回根节点
		}
		else//如果两个节点都在右子树，则递归处理右子树
		{
			return GetCommentParent(pRoot->m_nRight,pNode1,pNode2);
		}
	}
}

bool GetNodePath(BTree* pRoot, BTree* pNode, list<BTree*> & path)
{
	if (pRoot == pNode)
	{
		path.push_back(pRoot);
		return true;
	}
	if (pRoot == NULL)
	{
		return false; 
	}

	path.push_back(pRoot);
	bool found = false;
	found = GetNodePath(pRoot->m_nLeft,pNode,path);
	if (!found)
	{
		found = GetNodePath(pRoot->m_nRight,pNode,path);
	}
	if (!found)
	{
		path.pop_back();
	}
	return found;
}

BTree* GetCommentParent2(BTree* pRoot,BTree* pNode1,BTree* pNode2)
{
	if (pRoot == NULL || pNode1 == NULL || pNode2 == NULL)
	{
		return NULL;
	}
	list<BTree*> path1;
	list<BTree*> path2;
	bool bResult1 = GetNodePath(pRoot,pNode1,path1);
	bool bResult2 = GetNodePath(pRoot,pNode2,path2);
	if (!bResult1 || !bResult2)
	{
		return NULL;
	}

	list<BTree*>::const_iterator iter1 = path1.begin();
	list<BTree*>::const_iterator iter2 = path2.begin();
	BTree* pCommentParent = NULL;
	while (iter1 != path1.end() && iter2 != path2.end())
	{
		if (*iter1 == *iter2)
		{
			pCommentParent = *iter1;
		}
		
		iter1++;
		iter2++;
	}

	return pCommentParent;
}

BTree*  findOneNode(BTree* pRoot, int value)
{
	if (pRoot == NULL)
	{
		return NULL;
	}
	queue<BTree*> q;
	q.push(pRoot);
	BTree* ret = NULL;
	while (!q.empty())
	{
		BTree* pNode = q.front();
		q.pop();
		if (pNode->m_nValue == value)
		{
			return pNode;
		}
		else
		{
			if (pNode->m_nLeft != NULL)
			{
				q.push(pNode->m_nLeft);
			}
			if (pNode->m_nRight != NULL)
			{
				q.push(pNode->m_nRight);
			}
		}
	}
}

int _tmain(int argc, _TCHAR* argv[])
{
	BTree* m_pRoot = new BTree(5);
	Insert(6,m_pRoot);
	Insert(3,m_pRoot);
	Insert(4,m_pRoot);
	Insert(2,m_pRoot);

	BTree* node = findOneNode(m_pRoot,2);
	BTree* node2 = findOneNode(m_pRoot,6);

	BTree* pNode = GetCommentParent2(m_pRoot,node,node2);
	cout<<pNode->m_nValue<<endl;
	getchar();
	return 0;
}

C++算法之 求二叉树两个节点的最低公共节点

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原文地址：http://blog.csdn.net/djb100316878/article/details/42672193