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把一个有序链表构成成平衡二叉树。和上一题有一点像。
思路一：将有序链表存在一个数组里。然后根据每次访问中间节点当做根节点递归左右子树节点即可。代码如下：
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode *sortedArrayTree(vector<int> arr, int start, int end) { if (start > end) return NULL; TreeNode *root = new TreeNode(arr[(start + end)/2]); root -> left = sortedArrayTree(arr, start, (start + end)/2 - 1); root -> right = sortedArrayTree(arr, (start + end)/2 + 1, end); return root; } // 给定有序链表，构造高度平衡二叉树 TreeNode *sortedListToBST(ListNode *head) { if (!head) return NULL; vector<int> tmp; while(head) { tmp.push_back(head -> val); head = head -> next; } return sortedArrayTree(tmp, 0, tmp.size() - 1); } };
思路和做法应该是对的，但是Memory Limit Exceeded了，说明不能用数组存，没有那么大的空间，那就之间在链表上操作。是否记得我们在Construct Binary Tree from Inorder and Postorder Traversal中也遇到过Memory Limit的问题。那里也是应为开辟的空间有点大了。
其实巧妙的是，我发现如果我们这里把传入的arr当做应用传入，也就是vector<int> &arr的话，就可以Accept。不信你改改试试。不过我们还是再想想，直接链表上怎么操作吧。
这个是用两个链表节点递归的，节点相同就返回null，找中间节点用代码中的while操作，java代码如下：
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; next = null; } * } */ /** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public TreeNode sortedListToBST(ListNode head) { return rec(head, null); } public TreeNode rec(ListNode start, ListNode end) { if(start == end) { return null; } ListNode p = start, q = start; while(q != end && q.next != end) { p = p.next; q = q.next.next; } TreeNode root = new TreeNode(p.val); root.left = rec(start, p); root.right = rec(p.next, end); return root; } }
这个是用传入长度，然后找到中间节点：
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode *sortedListToBST(ListNode *head) { int n=0; ListNode *p=head; while(p!=NULL)n++,p=p->next; return build(head,n); } TreeNode *build(ListNode *head,int n) { if(head==NULL||n==0)return NULL; ListNode *p=head; for(int i=1;i<(n+1)/2;++i)p=p->next; TreeNode *root=new TreeNode(p->val); root->left=build(head,(n+1)/2-1); root->right=build(p->next,n-(n+1)/2); } };
leetcode上讨论组的最佳解法是自底向上的：
BinaryTree* sortedListToBST(ListNode *& list, int start, int end) { if (start > end) return NULL; // same as (start+end)/2, avoids overflow int mid = start + (end - start) / 2; BinaryTree *leftChild = sortedListToBST(list, start, mid-1); BinaryTree *parent = new BinaryTree(list->data); parent->left = leftChild; list = list->next; parent->right = sortedListToBST(list, mid+1, end); return parent; } BinaryTree* sortedListToBST(ListNode *head, int n) { return sortedListToBST(head, 0, n-1); }
leetcode Convert Sorted List to Binary Search Tree
标签：style blog http io ar color os sp java
原文地址：http://www.cnblogs.com/higerzhang/p/4132072.html
