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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:
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] <= 3000preorderandinorderconsist of unique values.- Each value of
inorderalso appears inpreorder. preorderis guaranteed to be the preorder traversal of the tree.inorderis guaranteed to be the inorder traversal of the tree.
# time complexity: O(n)
# space complexity: O(n)
from typing import List, Optional
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
def dfs(left: int, right: int):
nonlocal preorderIndex
if left > right:
return None
rootValue = preorder[preorderIndex]
root = TreeNode(rootValue)
preorderIndex += 1
root.left = dfs(left, inorderMap[rootValue] - 1)
root.right = dfs(inorderMap[rootValue] + 1, right)
return root
preorderIndex = 0
inorderMap = {}
for i, value in enumerate(inorder):
inorderMap[value] = i
return dfs(0, len(inorder) - 1)
preorder = [3, 9, 20, 15, 7]
inorder = [9, 3, 15, 20, 7]
print(Solution().buildTree(preorder, inorder))
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