Update 0108.将有序数组转换为二叉搜索树.md

This commit is contained in:
jianghongcheng
2023-05-23 21:46:41 -05:00
committed by GitHub
parent 53652744d2
commit 76e3811c5e

View File

@ -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)]
def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
if len(nums) == 0:
return None
while nodeSt:
node = nodeSt.pop() # 处理根节点
left = leftSt.pop()
right = rightSt.pop()
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
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)
curNode.val = nums[mid] # 将mid对应的元素给中间节点
if right > mid + 1: # 处理区间
node.right = TreeNode()
nodeSt.append(node.right)
leftSt.append(mid + 1)
rightSt.append(right)
if left <= mid - 1: # 处理区间
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