diff --git a/problems/0108.将有序数组转换为二叉搜索树.md b/problems/0108.将有序数组转换为二叉搜索树.md index 89778421..056ef3e2 100644 --- a/problems/0108.将有序数组转换为二叉搜索树.md +++ b/problems/0108.将有序数组转换为二叉搜索树.md @@ -316,73 +316,65 @@ class Solution { ``` ## Python -**递归** - +递归法 ```python -# 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]) -> TreeNode: - ''' - 构造二叉树:重点是选取数组最中间元素为分割点,左侧是递归左区间;右侧是递归右区间 - 必然是平衡树 - 左闭右闭区间 - ''' - # 返回根节点 - root = self.traversal(nums, 0, len(nums)-1) - return root - def traversal(self, nums: List[int], left: int, right: int) -> TreeNode: - # Base Case if left > right: return None - # 确定左右界的中心,防越界 mid = left + (right - left) // 2 - # 构建根节点 - mid_root = TreeNode(nums[mid]) - # 构建以左右界的中心为分割点的左右子树 - mid_root.left = self.traversal(nums, left, mid-1) - mid_root.right = self.traversal(nums, mid+1, right) + root = TreeNode(nums[mid]) + root.left = self.traversal(nums, left, mid - 1) + root.right = self.traversal(nums, mid + 1, right) + return root + + def sortedArrayToBST(self, nums: List[int]) -> TreeNode: + root = self.traversal(nums, 0, len(nums) - 1) + return root - # 返回由被传入的左右界定义的某子树的根节点 - return mid_root ``` -**迭代**(左闭右开) +迭代法 ```python +from collections import deque + class Solution: - def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: - if len(nums) == 0: return None - root = TreeNode() # 初始化 - nodeSt = [root] - leftSt = [0] - rightSt = [len(nums)] - - while nodeSt: - node = nodeSt.pop() # 处理根节点 - left = leftSt.pop() - right = rightSt.pop() - mid = left + (right - left) // 2 - node.val = nums[mid] - - if left < mid: # 处理左区间 - node.left = TreeNode() - nodeSt.append(node.left) - leftSt.append(left) - rightSt.append(mid) - - if right > mid + 1: # 处理右区间 - node.right = TreeNode() - nodeSt.append(node.right) - leftSt.append(mid + 1) - rightSt.append(right) + def sortedArrayToBST(self, nums: List[int]) -> TreeNode: + if len(nums) == 0: + return None + root = TreeNode(0) # 初始根节点 + nodeQue = deque() # 放遍历的节点 + leftQue = deque() # 保存左区间下标 + rightQue = deque() # 保存右区间下标 + + nodeQue.append(root) # 根节点入队列 + leftQue.append(0) # 0为左区间下标初始位置 + rightQue.append(len(nums) - 1) # len(nums) - 1为右区间下标初始位置 + + while nodeQue: + curNode = nodeQue.popleft() + left = leftQue.popleft() + right = rightQue.popleft() + mid = left + (right - left) // 2 + + curNode.val = nums[mid] # 将mid对应的元素给中间节点 + + if left <= mid - 1: # 处理左区间 + curNode.left = TreeNode(0) + nodeQue.append(curNode.left) + leftQue.append(left) + rightQue.append(mid - 1) + + if right >= mid + 1: # 处理右区间 + curNode.right = TreeNode(0) + nodeQue.append(curNode.right) + leftQue.append(mid + 1) + rightQue.append(right) + return root + ``` ## Go