Question

109. 有序链表转换二叉搜索树

English Version

题目描述

给定一个单链表的头节点  head ，其中的元素 按升序排序 ，将其转换为 平衡 二叉搜索树。

 

示例 1:

输入: head = [-10,-3,0,5,9]
输出: [0,-3,9,-10,null,5]
解释: 一个可能的答案是[0，-3,9，-10,null,5]，它表示所示的高度平衡的二叉搜索树。

示例 2:

输入: head = []
输出: []

 

提示:

• head 中的节点数在[0, 2 * 104] 范围内
• -105 <= Node.val <= 105

解法

方法一

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def sortedListToBST(self, head: ListNode) -> TreeNode:
    def buildBST(nums, start, end):
      if start > end:
        return None
      mid = (start + end) >> 1
      return TreeNode(
        nums[mid], buildBST(nums, start, mid - 1), buildBST(nums, mid + 1, end)
      )

    nums = []
    while head:
      nums.append(head.val)
      head = head.next
    return buildBST(nums, 0, len(nums) - 1)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode() {}
 *   ListNode(int val) { this.val = val; }
 *   ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public TreeNode sortedListToBST(ListNode head) {
    List<Integer> nums = new ArrayList<>();
    for (; head != null; head = head.next) {
      nums.add(head.val);
    }
    return buildBST(nums, 0, nums.size() - 1);
  }

  private TreeNode buildBST(List<Integer> nums, int start, int end) {
    if (start > end) {
      return null;
    }
    int mid = (start + end) >> 1;
    TreeNode root = new TreeNode(nums.get(mid));
    root.left = buildBST(nums, start, mid - 1);
    root.right = buildBST(nums, mid + 1, end);
    return root;
  }
}

/**
 * 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:
  TreeNode* sortedListToBST(ListNode* head) {
    vector<int> nums;
    for (; head != nullptr; head = head->next) {
      nums.push_back(head->val);
    }
    return buildBST(nums, 0, nums.size() - 1);
  }

private:
  TreeNode* buildBST(vector<int>& nums, int start, int end) {
    if (start > end) {
      return nullptr;
    }
    int mid = (start + end) / 2;
    TreeNode* root = new TreeNode(nums[mid]);
    root->left = buildBST(nums, start, mid - 1);
    root->right = buildBST(nums, mid + 1, end);
    return root;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func sortedListToBST(head *ListNode) *TreeNode {
  nums := []int{}
  for head != nil {
    nums = append(nums, head.Val)
    head = head.Next
  }
  return buildBST(nums, 0, len(nums)-1)
}

func buildBST(nums []int, start, end int) *TreeNode {
  if start > end {
    return nil
  }
  mid := (start + end) >> 1
  return &TreeNode{
    Val:   nums[mid],
    Left:  buildBST(nums, start, mid-1),
    Right: buildBST(nums, mid+1, end),
  }
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

const find = (start: ListNode | null, end: ListNode | null) => {
  let fast = start;
  let slow = start;
  while (fast !== end && fast.next !== end) {
    fast = fast.next.next;
    slow = slow.next;
  }
  return slow;
};

const build = (start: ListNode | null, end: ListNode | null) => {
  if (start == end) {
    return null;
  }
  const node = find(start, end);
  return new TreeNode(node.val, build(start, node), build(node.next, end));
};

function sortedListToBST(head: ListNode | null): TreeNode | null {
  return build(head, null);
}

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//   ListNode {
//     next: None,
//     val
//   }
//   }
// }
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn build(vals: &Vec<i32>, start: usize, end: usize) -> Option<Rc<RefCell<TreeNode>>> {
    if start == end {
      return None;
    }
    let mid = (start + end) >> 1;
    Some(
      Rc::new(
        RefCell::new(TreeNode {
          val: vals[mid],
          left: Self::build(vals, start, mid),
          right: Self::build(vals, mid + 1, end),
        })
      )
    )
  }

  pub fn sorted_list_to_bst(head: Option<Box<ListNode>>) -> Option<Rc<RefCell<TreeNode>>> {
    let mut vals = Vec::new();
    let mut cur = &head;
    while let Some(node) = cur {
      vals.push(node.val);
      cur = &node.next;
    }
    Self::build(&vals, 0, vals.len())
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.next = (next===undefined ? null : next)
 * }
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {ListNode} head
 * @return {TreeNode}
 */
var sortedListToBST = function (head) {
  const buildBST = (nums, start, end) => {
    if (start > end) {
      return null;
    }
    const mid = (start + end) >> 1;
    const root = new TreeNode(nums[mid]);
    root.left = buildBST(nums, start, mid - 1);
    root.right = buildBST(nums, mid + 1, end);
    return root;
  };

  const nums = new Array();
  for (; head != null; head = head.next) {
    nums.push(head.val);
  }
  return buildBST(nums, 0, nums.length - 1);
};

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   struct ListNode *next;
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   struct TreeNode *left;
 *   struct TreeNode *right;
 * };
 */
struct ListNode* find(struct ListNode* start, struct ListNode* end) {
  struct ListNode* fast = start;
  struct ListNode* slow = start;
  while (fast != end && fast->next != end) {
    fast = fast->next->next;
    slow = slow->next;
  }
  return slow;
}

struct TreeNode* bulid(struct ListNode* start, struct ListNode* end) {
  if (start == end) {
    return NULL;
  }
  struct ListNode* node = find(start, end);
  struct TreeNode* ans = malloc(sizeof(struct TreeNode));
  ans->val = node->val;
  ans->left = bulid(start, node);
  ans->right = bulid(node->next, end);
  return ans;
}

struct TreeNode* sortedListToBST(struct ListNode* head) {
  return bulid(head, NULL);
}