Reorder List and Sum of Left Leaves

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes’ values.

For example,

Given {1,2,3,4}, reorder it to {1,4,2,3}.

这个题目有点类似于综合题的感觉。对比给出实例。可以看出是将后半部的链表翻转后再与前半部合并。

算法分析：

1. 首先要分解链表，将链表一分为二。

2. 将后半部分链表倒序。

3. 合并两个链表。

class  Solution {
public:
void reoderList(ListNode* head){
if(head == NULL)
return ;
ListNode*   p1 = head;
ListNode*    p2 = splitList(head);

p2 = reverseList(p2);
mergelist(p1,p2);
}

ListNode*  splitList(ListNode* head){
ListNode*   slow = new ListNode(0);
slow->next = head;
ListNode*   fast = slow;
while(fast->next && fast->next->next){
slow = slow->next;
fast = fast->next->next;
}

if(fast->next){
slow = slow->next;
fast = fast->next;
}
ListNode*  tmp = slow->next;
slow->next =NULL;
return tmp;
}

ListNode*  reverseList(ListNode*  head){
if(head == NULL)
return head;
ListNode*  p=head;
p = head->next;
head->next = NULL;
while(p){
ListNode*   tmp = p->next;
p->next = head;
head = p;
p = tmp;
}
return head;
}

void  mergelist(ListNode* p1,ListNode*  p2){
while(p2){
ListNode*   tmp = p2;
p2 = p2->next;
p1->next =p2;
tmp->next = p1->next;
p1 = p1->next->next;
}
}
};

再来看看二叉树：Sum of Left Leaves

计算所有左子树之和。

这个很容易理解，进行递归求解：

int  sumOfLeftLeaves(ListNode* root){
if(root == NULL)  return 0;
if(root->left =NULL && root->left->left =NULL
&& root->left->right == NULL){
return root->val + sumOfLeftLeaves(root->right);
}
return sumOfLeftLeaves(root->left) +
sumOfLeftLeaves(root->right);

}

好了，今天的解析到此为止了。