Question

面试题 04.02. 最小高度树

English Version

题目描述

给定一个有序整数数组，元素各不相同且按升序排列，编写一个算法，创建一棵高度最小的二叉搜索树。

示例:

给定有序数组: [-10,-3,0,5,9],

一个可能的答案是：[0,-3,9,-10,null,5]，它可以表示下面这个高度平衡二叉搜索树：

      0 
     / \ 
     -3   9 
     /   / 
   -10  5 

解法

方法一：递归

先找到数组的中间点，作为二叉搜索树的根节点，然后递归左右子树即可。

时间复杂度 $O(n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组长度。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
    def dfs(l, r):
      if l > r:
        return None
      mid = (l + r) >> 1
      return TreeNode(nums[mid], dfs(l, mid - 1), dfs(mid + 1, r))

    return dfs(0, len(nums) - 1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  private int[] nums;

  public TreeNode sortedArrayToBST(int[] nums) {
    this.nums = nums;
    return dfs(0, nums.length - 1);
  }

  private TreeNode dfs(int l, int r) {
    if (l > r) {
      return null;
    }
    int mid = (l + r) >> 1;
    return new TreeNode(nums[mid], dfs(l, mid - 1), dfs(mid + 1, r));
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  TreeNode* sortedArrayToBST(vector<int>& nums) {
    function<TreeNode*(int, int)> dfs = [&](int l, int r) -> TreeNode* {
      if (l > r) return nullptr;
      int mid = l + r >> 1;
      return new TreeNode(nums[mid], dfs(l, mid - 1), dfs(mid + 1, r));
    };
    return dfs(0, nums.size() - 1);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func sortedArrayToBST(nums []int) *TreeNode {
  var dfs func(int, int) *TreeNode
  dfs = func(l, r int) *TreeNode {
    if l > r {
      return nil
    }
    mid := (l + r) >> 1
    return &TreeNode{nums[mid], dfs(l, mid-1), dfs(mid+1, r)}
  }

  return dfs(0, len(nums)-1)
}

/**
 * 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)
 *   }
 * }
 */

function sortedArrayToBST(nums: number[]): TreeNode | null {
  const dfs = (l: number, r: number): TreeNode | null => {
    if (l > r) {
      return null;
    }
    const mid = (l + r) >> 1;
    return new TreeNode(nums[mid], dfs(l, mid - 1), dfs(mid + 1, r));
  };
  return dfs(0, nums.length - 1);
}

// 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 dfs(nums: &Vec<i32>, start: usize, end: usize) -> Option<Rc<RefCell<TreeNode>>> {
    if start >= end {
      return None;
    }
    let mid = start + (end - start) / 2;
    Some(
      Rc::new(
        RefCell::new(TreeNode {
          val: nums[mid],
          left: Self::dfs(nums, start, mid),
          right: Self::dfs(nums, mid + 1, end),
        })
      )
    )
  }
  pub fn sorted_array_to_bst(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
    let end = nums.len();
    Self::dfs(&nums, 0, end)
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *   this.val = val;
 *   this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} nums
 * @return {TreeNode}
 */
var sortedArrayToBST = function (nums) {
  function dfs(l, r) {
    if (l > r) {
      return null;
    }
    const mid = (l + r) >> 1;
    return new TreeNode(nums[mid], dfs(l, mid - 1), dfs(mid + 1, r));
  }

  return dfs(0, nums.length - 1);
};