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今天做了一题求二叉树节点的最大距离,顺便写了下二叉树的建立,遍历的过程。
我觉得这题的主要思想是深度遍历+动态规划,我们在深度遍历的过程中,对于某一个子树,求出左右子树叶子节点到根节点的最大距离,进而求出经过根节点的最大距离。 最后求出所有子树经过根节点的最大距离。就是这个题目的最终结果。代码如下:
//二叉树的建立,以及遍历
//16 14 8 2 -1 -1 4 -1 -1 7 1 -1 -1 -1 10 9 -1 -1 3 -1 -1
//16 14 8 2 -1 -1 4 -1 -1 7 1 -1 -1 -1 -1
void BuildBTree(BTreeNode* &pRoot)
{
int nTemp;
cin >> nTemp;
if (nTemp == -1)
{
pRoot = NULL;
}
else
{
pRoot = new BTreeNode();
pRoot->nValue = nTemp;
BuildBTree(pRoot->pLeft);
BuildBTree(pRoot->pRight);
}
}
void PreOrderBTree(BTreeNode* pRoot)
{
if (pRoot == NULL)
{
return;
}
cout << pRoot->nValue << " ";
PreOrderBTree(pRoot->pLeft);
PreOrderBTree(pRoot->pRight);
}
void MidOrderBTree(BTreeNode* pRoot)
{
if (pRoot == NULL)
{
return;
}
MidOrderBTree(pRoot->pLeft);
cout << pRoot->nValue << " ";
MidOrderBTree(pRoot->pRight);
}
void PostOrderBTree(BTreeNode* pRoot)
{
if (pRoot == NULL)
{
return;
}
PostOrderBTree(pRoot->pLeft);
PostOrderBTree(pRoot->pRight);
cout << pRoot->nValue << " ";
}
int GetMaxNodeMaxDistance(BTreeNode* pRoot)
{
if (pRoot == NULL)
{
return -1;
}
//左子树叶子结点到根结点的最大距离
int max_left_distance = GetMaxNodeMaxDistance(pRoot->pLeft);
//右子树叶子结点到根结点的最大距离
int max_right_distance = GetMaxNodeMaxDistance(pRoot->pRight);
//每个子树节点的最大距离
int max_root_distance = max_left_distance + max_right_distance + 2;
//比较每个子树节点的最大距离
if (max_root_distance > max_distance)
{
max_distance = max_root_distance;
}
return max_left_distance > max_right_distance ? max_left_distance+1 : max_right_distance+1;
}
int max_distance = 0;
int main()
{
BTreeNode* Root = NULL;
BuildBTree(Root);
cout << "----------------Build End------------------" << endl;
system("pause");
cout << "PreOrderBTree:" << endl;
PreOrderBTree(Root);
cout << endl << "MidOrderBTree:" << endl;
MidOrderBTree(Root);
cout << endl << "PostOrderBTree:" << endl;
PostOrderBTree(Root);
cout << endl << "GetMaxNodeMaxDistance:" << endl;
GetMaxNodeMaxDistance(Root);
cout << max_distance << endl;
system("pause");
return 0;
}版权声明:本文为博主原创文章,未经博主允许不得转载。
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原文地址:http://blog.csdn.net/king__moving/article/details/48025457