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求一个二叉树中任意两个节点间的最大距离,两个节点的距离的定义

题目: 求一个二叉树中任意两个节点间的最大距离,两个节点的距离的定义是这两个节点间边的个数,比如某个孩子节点和父节点间的距离是1,和相邻兄弟节点间的距离是2, 优化时间空间杂度。 思路一: 计算一个二叉树的最大距离有两个情况: 情况A: 路径经过左子

题目:
求一个二叉树中任意两个节点间的最大距离,两个节点的距离的定义是这两个节点间边的个数,比如某个孩子节点和父节点间的距离是1,和相邻兄弟节点间的距离是2,

优化时间空间杂度。

思路一:

计算一个二叉树的最大距离有两个情况:
情况A: 路径经过左子树的最深节点,通过根节点,再到右子树的最深节点。
情况B: 路径不穿过根节点,而是左子树或右子树的最大距离路径,取其大者。
首先算出经过根节点的最大路径的距离,其实就是左右子树的深度和;然后分别算出左子树和右子树的最大距离,三者比较,最大值就是当前二叉树的最大距离了。

代码如下:

[cpp] view plaincopyprint?

/*----------------------------- Copyright by yuucyf. 2011.09.02 ------------------------------*/ #include "stdafx.h" #include #include using namespace std; typedef struct tagSBTreeNode { tagSBTreeNode *psLeft; tagSBTreeNode *psRight; int nValue; int nMaxLeft; int nMaxRight; tagSBTreeNode() { psLeft = psRight = NULL; nValue = 0; nMaxLeft = nMaxRight = 0; } }S_TreeNode; void AddTreeNode(S_TreeNode *&psTreeNode, int nValue) { if (NULL == psTreeNode) { psTreeNode = new S_TreeNode; assert(NULL != psTreeNode); psTreeNode->nValue = nValue; } else if (psTreeNode->nValue { AddTreeNode(psTreeNode->psRight, nValue); } else AddTreeNode(psTreeNode->psLeft, nValue); } int MaxDepth( const S_TreeNode *psTreeNode) { int nDepth = 0; if (NULL != psTreeNode) { int nLeftDepth = MaxDepth(psTreeNode->psLeft); int nRightDepth = MaxDepth(psTreeNode->psRight); nDepth = (nLeftDepth > nRightDepth) ? nLeftDepth : nRightDepth; nDepth++; } return nDepth; } int MaxDistance( const S_TreeNode *psRootNode) { int nDistance = 0; if (NULL != psRootNode) { nDistance = MaxDepth(psRootNode->psLeft) + MaxDepth(psRootNode->psRight); int nLeftDistance = MaxDistance(psRootNode->psLeft); int nRightDistance= MaxDistance(psRootNode->psRight); nDistance = (nLeftDistance > nDistance) ? nLeftDistance : nDistance; nDistance = (nRightDistance > nDistance) ? nRightDistance : nDistance; } return nDistance; } int _tmain( int argc, _TCHAR* argv[]) { S_TreeNode *psRoot = NULL; AddTreeNode(psRoot, 9); AddTreeNode(psRoot, 6); AddTreeNode(psRoot, 4); AddTreeNode(psRoot, 8); AddTreeNode(psRoot, 7); AddTreeNode(psRoot, 15); AddTreeNode(psRoot, 13); AddTreeNode(psRoot, 16); AddTreeNode(psRoot, 18); cout "任意两个节点间的最大距离为:" return 0; }

/*-----------------------------
Copyright by yuucyf. 2011.09.02
------------------------------*/
#include "stdafx.h"
#include  
#include 
using namespace std;

typedef struct tagSBTreeNode
{
	tagSBTreeNode *psLeft;
	tagSBTreeNode *psRight;
	int	nValue;

	int nMaxLeft;
	int nMaxRight;

	tagSBTreeNode()
	{
		psLeft = psRight = NULL;
		nValue = 0;
		nMaxLeft = nMaxRight = 0;
	}
}S_TreeNode;


void AddTreeNode(S_TreeNode *&psTreeNode, int nValue)
{
	if (NULL == psTreeNode)
	{
		psTreeNode = new S_TreeNode;
		assert(NULL != psTreeNode);
		psTreeNode->nValue = nValue;
	}
	else if (psTreeNode->nValue  psRight, nValue);
	}
	else
		AddTreeNode(psTreeNode->psLeft, nValue);
}

int MaxDepth(const S_TreeNode *psTreeNode)
{
	int nDepth = 0;
	if (NULL != psTreeNode)
	{
		int nLeftDepth = MaxDepth(psTreeNode->psLeft);
		int nRightDepth = MaxDepth(psTreeNode->psRight);
		nDepth = (nLeftDepth > nRightDepth) ? nLeftDepth : nRightDepth;
		nDepth++;
	}

	return nDepth;
}

int MaxDistance(const S_TreeNode *psRootNode)
{
	int nDistance = 0;
	if (NULL != psRootNode)
	{
		nDistance = MaxDepth(psRootNode->psLeft) + MaxDepth(psRootNode->psRight);
		int nLeftDistance = MaxDistance(psRootNode->psLeft);
		int nRightDistance= MaxDistance(psRootNode->psRight);
		
		nDistance = (nLeftDistance > nDistance) ? nLeftDistance : nDistance;
		nDistance = (nRightDistance > nDistance) ? nRightDistance : nDistance;
	}
	
	return nDistance;
}




int _tmain(int argc, _TCHAR* argv[])
{
	S_TreeNode *psRoot = NULL;
	AddTreeNode(psRoot, 9);
	AddTreeNode(psRoot, 6);
	AddTreeNode(psRoot, 4);
	AddTreeNode(psRoot, 8);
	AddTreeNode(psRoot, 7);
	AddTreeNode(psRoot, 15);
	AddTreeNode(psRoot, 13);
	AddTreeNode(psRoot, 16);
	AddTreeNode(psRoot, 18);

	cout  


思路二:

思路一不是效率最高的,因为在计算二叉树的深度的时候存在重复计算。但应该是可读性比较好的,同时也没有改变原有二叉树的结构和使用额外的全局变量。这里之间给出代码,因为代码的注释已经写的非常详细了。

代码如下:

[cpp] view plaincopyprint?

int g_nMaxLeft = 0; void MaxDistance_2(S_TreeNode *psRoot) { // 遍历到叶子节点,返回 if (NULL == psRoot) return ; // 如果左子树为空,那么该节点的左边最长距离为0 if (psRoot->psLeft == NULL) { psRoot->nMaxLeft = 0; } // 如果右子树为空,那么该节点的右边最长距离为0 if (psRoot->psRight == NULL) { psRoot -> nMaxRight = 0; } // 如果左子树不为空,递归寻找左子树最长距离 if (psRoot->psLeft != NULL) { MaxDistance_2(psRoot->psLeft); } // 如果右子树不为空,递归寻找右子树最长距离 if (psRoot->psRight != NULL) { MaxDistance_2(psRoot->psRight); } // 计算左子树最长节点距离 if (psRoot->psLeft != NULL) { int nTempMax = 0; if (psRoot->psLeft->nMaxLeft > psRoot->psLeft->nMaxRight) { nTempMax = psRoot->psLeft->nMaxLeft; } else { nTempMax = psRoot->psLeft->nMaxRight; } psRoot->nMaxLeft = nTempMax + 1; } // 计算右子树最长节点距离 if (psRoot->psRight != NULL) { int nTempMax = 0; if (psRoot->psRight->nMaxLeft > psRoot->psRight->nMaxRight) { nTempMax = psRoot->psRight->nMaxLeft; } else { nTempMax = psRoot->psRight->nMaxRight; } psRoot->nMaxRight = nTempMax + 1; } // 更新最长距离 if (psRoot->nMaxLeft + psRoot->nMaxRight > g_nMaxLeft) { g_nMaxLeft = psRoot->nMaxLeft + psRoot->nMaxRight; } }

int g_nMaxLeft = 0;
void MaxDistance_2(S_TreeNode *psRoot)
{
	// 遍历到叶子节点,返回
	if (NULL == psRoot)
		return;

	// 如果左子树为空,那么该节点的左边最长距离为0
	if (psRoot->psLeft == NULL)
	{
		psRoot->nMaxLeft = 0;
	}

	// 如果右子树为空,那么该节点的右边最长距离为0
	if (psRoot->psRight == NULL)
	{
		psRoot -> nMaxRight = 0;
	}

	// 如果左子树不为空,递归寻找左子树最长距离
	if (psRoot->psLeft != NULL)
	{
		MaxDistance_2(psRoot->psLeft);
	}

	// 如果右子树不为空,递归寻找右子树最长距离
	if (psRoot->psRight != NULL)
	{
		MaxDistance_2(psRoot->psRight);
	}

	// 计算左子树最长节点距离
	if (psRoot->psLeft != NULL)
	{
		int nTempMax = 0;
		if (psRoot->psLeft->nMaxLeft > psRoot->psLeft->nMaxRight)
		{
			nTempMax = psRoot->psLeft->nMaxLeft;
		}
		else
		{
			nTempMax = psRoot->psLeft->nMaxRight;
		}
		psRoot->nMaxLeft = nTempMax + 1;
	}

	// 计算右子树最长节点距离
	if (psRoot->psRight != NULL)
	{
		int nTempMax = 0;
		if(psRoot->psRight->nMaxLeft > psRoot->psRight->nMaxRight)
		{
			nTempMax = psRoot->psRight->nMaxLeft;
		}
		else
		{
			nTempMax = psRoot->psRight->nMaxRight;
		}
		psRoot->nMaxRight = nTempMax + 1;
	}

	// 更新最长距离
	if (psRoot->nMaxLeft + psRoot->nMaxRight > g_nMaxLeft)
	{
		g_nMaxLeft = psRoot->nMaxLeft + psRoot->nMaxRight;
	}
} 

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