有序链表

0->1->2->3->4->5

转换为一个二叉排序树。我们在此创建一个平衡二叉排序树

1.先找链表到中间的节点

2.中间节点的val创建一个新的树节点TreeNode

3.将链表断裂为2部分

4.递归创建左子树和右子树

#include<iostream>
#include<cstdlib>
using namespace std; struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}; class Solution {
public:
TreeNode* sortedListToBST(ListNode* head) {
//如果节点为空直接返回空指针
if(head==NULL){
return NULL;
}
//创建新节点,返回
if(head->next==NULL){
TreeNode * root;
root=new TreeNode(head->val);
return root;
}
//只有2个节点,取第一个作为根节点,第2个作为右节点
if(head->next->next==NULL){
TreeNode * root;
root=new TreeNode(head->val);
TreeNode * right;
right=new TreeNode(head->next->val);
root->right=right;
return root;
}
//找到中间节点
ListNode *step1,*step2;
step1=head;
step2=head;
ListNode *temp;//保存中间节点的前一个节点
while(step2->next!=NULL&&step2->next->next!=NULL){
temp=step1;
step1=step1->next;
step2=step2->next->next;
}
temp->next=NULL;//将链表在中间断开
//创建新的根节点
TreeNode *root ;
root=new TreeNode(step1->val);
//递归创建左右子树
TreeNode *left,*right;
left=sortedListToBST(head);
right=sortedListToBST(step1->next);
root->left=left;
root->right=right;
return root;
}
};
class Travel{
public:
void travel(TreeNode * root){
if(root==NULL){
return ;
}
//cout<<root->val<<endl;//先序遍历
travel(root->left);
//cout<<root->val<<endl;//中序遍历
travel(root->right);
//cout<<root->val<<endl;//后续遍历
}
}; main(){
ListNode *head,*tail;
head=new ListNode();
tail=head;
for(int i=;i<;i++){
ListNode *temp;
temp=new ListNode(i);
tail->next=temp;
tail=tail->next;
}
Solution s;
TreeNode *root;
root=s.sortedListToBST(head);
Travel t;
t.travel(root); }

ps:送一个遍历小程序

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