16.合并两个排序的链表
/**************************************************************/
/* 题目描述
输入两个单调递增的链表,输出两个链表合成后的链表,当然我们需要合成后的链表满足单调不减规则。 */
/**************************************************************/
class Solution {
public:
ListNode * Merge(ListNode* pHead1, ListNode* pHead2)
{
if (pHead1 == NULL && pHead2 != NULL)
return pHead2;
if (pHead2 == NULL && pHead1 != NULL)
return pHead1;
if (pHead2 == NULL && pHead1 == NULL)
return NULL;
ListNode *pNewHead = NULL;
if (pHead1->val > pHead2->val)
{
pNewHead = pHead2;
pHead2 = pHead2->next;
}
else
{
pNewHead = pHead1;
pHead1 = pHead1->next;
}
ListNode *current = pNewHead;
while (pHead1 != NULL && pHead2 != NULL)
{
if (pHead1->val > pHead2->val)
{
current->next = pHead2;
pHead2 = pHead2->next;
current = current->next;
}
else
{
current->next = pHead1;
pHead1 = pHead1->next;
current = current->next;
}
}
if (pHead1 != NULL)
{
current->next = pHead1;
}
if (pHead2 != NULL)
current->next = pHead2;
return pNewHead;
}
};
17.树的子结构
/************************************************************************/
/* 题目描述
输入两棵二叉树A,B,判断B是不是A的子结构。
(ps:我们约定空树不是任意一个树的子结构) */
/************************************************************************/
bool isHasB(TreeNode* pRoot1, TreeNode* pRoot2)
{
if (pRoot2 == NULL)
return true; //pRoot2为空说明前面的都匹配了
if (pRoot1 == NULL)
return false; //pRoot1为空,但是pRoot2不为空,说明没有全匹配
if (pRoot1->val != pRoot2->val)
return false;
//继续判断下一个结点
return isHasB(pRoot1->left, pRoot2->left) && isHasB(pRoot1->right, pRoot2->right);
}
bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
{
bool result = false;
if (pRoot1 == NULL || pRoot2 == NULL)
return false;
if (pRoot1->val == pRoot2->val)
{
result = isHasB(pRoot1, pRoot2);
}
//不匹配,继续遍历
if (!result)
{
result = HasSubtree(pRoot1->left, pRoot2);
}
if (!result)
{
result = HasSubtree(pRoot1->right, pRoot2);
}
return result;
}
18.二叉树的镜像
/************************************************************************/
/* 题目描述
操作给定的二叉树,将其变换为源二叉树的镜像 */
/************************************************************************/
void Mirror(TreeNode *pRoot) {
if (pRoot == NULL)
return;
TreeNode *tmp = pRoot->left;
pRoot->left = pRoot->right;
pRoot->right = tmp;
Mirror(pRoot->left);
Mirror(pRoot->right);
}