LC.109. 有序链表转换二叉搜索树

LC.109. 有序链表转换二叉搜索树

思路：中序遍历。

1.转有序数组分治思想。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int>a;
    void fun(ListNode* head){
        while(head){
            a.push_back(head->val);
            head=head->next;
        }
    }
    TreeNode* build(int l,int r){
         if(l>r) return nullptr;
         int mid=(l+r)>>1;
         TreeNode* rt=new TreeNode(a[mid]);
         rt->left=build(l,mid-1);
         rt->right=build(mid+1,r);
         return rt;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        fun(head);
         return build(0,a.size()-1);
    }
};

2.头结点占位分治。

class Solution {
public:
    int getLength(ListNode* head) {
        int ret = 0;
        for (; head != nullptr; ++ret, head = head->next);
        return ret;
    }

    TreeNode* buildTree(ListNode*& head, int left, int right) {
        if (left > right) {
            return nullptr;
        }
        int mid = (left + right)>>1;
        TreeNode* root = new TreeNode();
        root->left = buildTree(head, left, mid - 1);
        root->val = head->val;
        head = head->next;
        root->right = buildTree(head, mid + 1, right);
        return root;
    }

    TreeNode* sortedListToBST(ListNode* head) {
        int length = getLength(head);
        return buildTree(head, 0, length - 1);
    }
};