Question

108. Convert Sorted Array to Binary Search Tree

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Description

Given an integer array nums where the elements are sorted in ascending order, convert _it to a __height-balanced_ _binary search tree_.

 

Example 1:

Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:

Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.

 

Constraints:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums is sorted in a strictly increasing order.

Solutions

Solution 1: Binary Search + Recursion

We design a recursive function $dfs(l, r)$, which indicates that the node values of the current binary search tree to be constructed are all within the index range $[l, r]$ of the array nums. This function returns the root node of the constructed binary search tree.

The execution process of the function $dfs(l, r)$ is as follows:

1. If $l > r$, it means the current array is empty, return null.
2. If $l \leq r$, take the element with the index $mid = \lfloor \frac{l + r}{2} \rfloor$ in the array as the root node of the current binary search tree, where $\lfloor x \rfloor$ represents rounding down $x$.
3. Recursively construct the left subtree of the current binary search tree, whose root node value is the element with the index $mid - 1$ in the array, and the node values of the left subtree are all within the index range $[l, mid - 1]$ of the array.
4. Recursively construct the right subtree of the current binary search tree, whose root node value is the element with the index $mid + 1$ in the array, and the node values of the right subtree are all within the index range $[mid + 1, r]$ of the array.
5. Return the root node of the current binary search tree.

The answer is the return value of the function $dfs(0, n - 1)$.

The time complexity is $O(n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array nums.

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

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

/**
 * 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 {
  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;
    TreeNode left = dfs(l, mid - 1);
    TreeNode right = dfs(mid + 1, r);
    return new TreeNode(nums[mid], left, right);
  }
}

/**
 * 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* sortedArrayToBST(vector<int>& nums) {
    function<TreeNode*(int, int)> dfs = [&](int l, int r) -> TreeNode* {
      if (l > r) {
        return nullptr;
      }
      int mid = (l + r) >> 1;
      auto left = dfs(l, mid - 1);
      auto right = dfs(mid + 1, r);
      return new TreeNode(nums[mid], left, right);
    };
    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
    left, right := dfs(l, mid-1), dfs(mid+1, r)
    return &TreeNode{nums[mid], left, right}
  }
  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 n = nums.length;
  if (n === 0) {
    return null;
  }
  const mid = n >> 1;
  return new TreeNode(
    nums[mid],
    sortedArrayToBST(nums.slice(0, mid)),
    sortedArrayToBST(nums.slice(mid + 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 to_bst(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::to_bst(nums, start, mid),
          right: Self::to_bst(nums, mid + 1, end),
        })
      )
    )
  }

  pub fn sorted_array_to_bst(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
    Self::to_bst(&nums, 0, nums.len())
  }
}

/**
 * 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 {number[]} nums
 * @return {TreeNode}
 */
var sortedArrayToBST = function (nums) {
  const dfs = (l, r) => {
    if (l > r) {
      return null;
    }
    const mid = (l + r) >> 1;
    const left = dfs(l, mid - 1);
    const right = dfs(mid + 1, r);
    return new TreeNode(nums[mid], left, right);
  };
  return dfs(0, nums.length - 1);
};