二叉樹中最遠的兩個結點的距離

解法一：求出每個結點作為根結點時的最遠距離（時間複雜度太高）

struct BinTreeNode
{
	int _data;
	BinTreeNode* _left;
	BinTreeNode* _right;
	BinTreeNode(int data = 0) :_data(data), _left(NULL), _right(NULL)
	{}
};

int DepthBin(BinTreeNode* root)
{
	if (root == NULL)
		return 0;
	int left = DepthBin(root->_left);
	int right = DepthBin(root->_right);
	return left > right ? left + 1 : right + 1;
}

void FindMaxLenOfBinTree(BinTreeNode* root, int& maxPath)
{
	if (root == NULL)
		return;
	int leftDepth = DepthBin(root->_left);
	int rightDepth = DepthBin(root->_right);
	if (maxPath < leftDepth + rightDepth)
	{
		maxPath = leftDepth + rightDepth;
	}
	FarestRoot(root->_left, maxPath);
	FarestRoot(root->_right, maxPath);
}

int FindMaxLen(BinTreeNode* root)
{
	int maxPath = 0;
	FarestRoot(root, maxPath);
	return maxPath;
}

解法二：程式碼註釋很詳細

struct BinTreeNode
{
	int _data;
	int _maxLeft;
	int _maxRight;
	BinTreeNode* _left;
	BinTreeNode* _right;
	BinTreeNode(int data = 0) 
		:_data(data), _maxLeft(0), _maxRight(0)
		,_left(NULL), _right(NULL)
	{}
};

void FindMaxLenOfBinTree(BinTreeNode* root, int& maxLen)
{
	if (root == NULL)
		return;

	//左子樹為空，該節點的左邊最長距離為0
	if (root->_left == NULL)
	{
		root->_maxLeft = 0;
	}
	//右子樹為空，該節點的右邊最長距離為0
	if (root->_right == NULL)
	{
		root->_maxRight = 0;
	}
	if (root->_left != NULL)
	{
		FindMaxLenOfBinTree(root->_left, maxLen);
	}
	if (root->_right != NULL)
	{
		FindMaxLenOfBinTree(root->_right, maxLen);
	}
	if (root->_left != NULL)
	{
		int tmp = 0;
		if (root->_left->_maxLeft > root->_left->_maxRight)
		{
			tmp = root->_left->_maxLeft;
		}
		else
		{
			tmp = root->_left->_maxRight;
		}
		root->_maxLeft = tmp + 1;
	}
	if (root->_right != NULL)
	{
		int tmp = 0;
		if (root->_right->_maxLeft > root->_right->_maxRight)
		{
			tmp = root->_right->_maxLeft;
		}
		else
		{
			tmp = root->_right->_maxRight;
		}
		root->_maxRight = tmp + 1;
	}
	if (root->_maxLeft + root->_maxRight > maxLen)
	{
		maxLen = root->_maxLeft + root->_maxRight;
	}
}

int FindMaxLen(BinTreeNode* root)
{
	int maxLen = 0;
	FindMaxLenOfBinTree(root, maxLen);
	return maxLen;
}