108. Convert Sorted Array to Binary Search Tree

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 \(\textit{dfs}(l, r)\), which represents that the values of the nodes to be constructed in the current binary search tree are within the index range \([l, r]\) of the array \(\textit{nums}\). This function returns the root node of the constructed binary search tree.

The execution process of the function \(\textit{dfs}(l, r)\) is as follows:

If \(l > r\), it means the current array is empty, so return null.
If \(l \leq r\), take the element at index \(\textit{mid} = \lfloor \frac{l + r}{2} \rfloor\) of the array as the root node of the current binary search tree, where \(\lfloor x \rfloor\) denotes the floor function of \(x\).
Recursively construct the left subtree of the current binary search tree, with the root node's value being the element at index \(\textit{mid} - 1\) of the array. The values of the nodes in the left subtree are within the index range \([l, \textit{mid} - 1]\) of the array.
Recursively construct the right subtree of the current binary search tree, with the root node's value being the element at index \(\textit{mid} + 1\) of the array. The values of the nodes in the right subtree are within the index range \([\textit{mid} + 1, r]\) of the array.
Return the root node of the current binary search tree.

The answer is the return value of the function \(\textit{dfs}(0, n - 1)\).

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

Python3JavaC++GoTypeScriptRustJavaScriptC#

# 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: int, r: int) -> Optional[TreeNode]:
            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() {}
 *     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;
        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() : 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) {
        auto dfs = [&](this auto&& 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 {
    pub fn sorted_array_to_bst(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        fn dfs(nums: &Vec<i32>, l: usize, r: usize) -> Option<Rc<RefCell<TreeNode>>> {
            if l > r {
                return None;
            }
            let mid = (l + r) / 2;
            if mid >= nums.len() {
                return None;
            }
            let mut node = Rc::new(RefCell::new(TreeNode::new(nums[mid])));
            node.borrow_mut().left = dfs(nums, l, mid - 1);
            node.borrow_mut().right = dfs(nums, mid + 1, r);
            Some(node)
        }
        dfs(&nums, 0, nums.len() - 1)
    }
}

/**
 * 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;
        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.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public 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));
    }
}

108. Convert Sorted Array to Binary Search Tree

Description

Solutions

Solution 1: Binary Search + Recursion

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