Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.
Example 1:
3
/ \
9 20
/ \
15 7
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1]
Output: [-1]
Constraints:
1 <= preorder.length <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorder and inorder consist of unique values.- Each value of
inorder also appears in preorder.
preorder is guaranteed to be the preorder traversal of the tree.inorder is guaranteed to be the inorder traversal of the tree.
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# 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 buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
n = len(preorder)
inorder_index_map = {}
for index, value in enumerate(inorder):
inorder_index_map[value] = index
def build(pre_l, pre_r, in_l, in_r) -> Optional[TreeNode]:
if pre_l > pre_r:
return None
root_val = preorder[pre_l]
root = TreeNode(root_val)
idx = inorder_index_map[root_val]
left_inorder_size = idx - in_l
root.left = build(
pre_l + 1,
pre_l + left_inorder_size,
in_l,
idx - 1
)
root.right = build(
pre_l + left_inorder_size + 1,
pre_r,
idx + 1,
in_r
)
return root
return build(0, n-1, 0, n-1)
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