105. 从前序与中序遍历序列构造二叉树
根据一棵树的前序遍历与中序遍历构造二叉树。
注意: 你可以假设树中没有重复的元素。
例如,给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树:
3
/ \
9 20
/ \
15 7
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal
Link:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
递归
O(N^2), 如果这个树是一个链表的话, N个节点,每个节点创建O(N)
基本的思路是用postorder最后一个作为跟节点,然后根据inorder值的位置,来确定左子树+右子树
普通递归
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if len(preorder) == 0 or len(inorder) == 0:
return None
val = preorder.pop(0)
root = TreeNode(val)
i = inorder.index(val)
root.left = self.buildTree(preorder, inorder[:i])
root.right = self.buildTree(preorder, inorder[i + 1:])
return root
时间优化递归
O(N)
⚠️注意,由于没有重复元素
- 先用字典做一个反向索引,加速查找index的速度
- 使用下标,而不是新建一个数组
- 反转preorder, 每次pop最后一个
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
self.map_inorder = {}
for i in range(len(inorder)):
self.map_inorder[inorder[i]] = i
preorder.reverse()
return self.helper(preorder, inorder, 0, len(inorder) - 1)
def helper(self, preorder: List[int], inorder: List[int], low: int, high: int) -> TreeNode:
if low > high:
return None
val = preorder.pop()
root = TreeNode(val)
i = self.map_inorder[val]
root.left = self.helper(preorder, inorder, low, i - 1)
root.right = self.helper(preorder, inorder, i + 1, high)
return root
空间优化递归
先创建左子树部分,然后在做右边
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
preorder.reverse()
inorder.reverse()
return self.helper(preorder, inorder, None)
def helper(self, preorder: List[int], inorder: List[int], stop: TreeNode) -> TreeNode:
if len(inorder) == 0 or inorder[-1] == stop:
return None
val = preorder.pop()
root = TreeNode(val)
root.left = self.helper(preorder, inorder, val)
inorder.pop()
root.right = self.helper(preorder, inorder, stop)
return root
–End–