leetcode 108. Convert Sorted Array to Binary Search Tree

最新推荐文章于 2025-06-04 15:25:41 发布

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最新推荐文章于 2025-06-04 15:25:41 发布

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本文介绍如何将一个按升序排列的数组转换为高度平衡的二叉搜索树。平衡二叉搜索树定义为对于每个节点，其两个子树的深度相差不超过1。文章提供了两种解决方案：递归方法和迭代方法。

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

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

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        def build_bst(arr, i, j):
            if i > j:
                return None
            mid = (i+j)>>1
            node = TreeNode(arr[mid])
            node.left = build_bst(arr, i, mid-1)
            node.right = build_bst(arr, mid+1, j)
            return node            
        
        return build_bst(nums, 0, len(nums)-1)

迭代解法，本质上是先序遍历：

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

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """        
        if not nums: return None
        q = [(0, len(nums)-1)]
        ans = TreeNode(0)
        nodes = [ans]
        while q:
            i, j = q.pop()
            mid = (i+j)>>1
            node = nodes.pop()
            node.val = nums[mid]
            if mid+1 <= j:
                node.right = TreeNode(0)
                q.append((mid+1, j))
                nodes.append(node.right)
            if mid-1 >= i:
                node.left = TreeNode(0)
                q.append((i, mid-1))
                nodes.append(node.left)
        return ans