Problem

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:

!https://assets.leetcode.com/uploads/2021/02/19/tree.jpg

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Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

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Input: preorder = [-1], inorder = [-1]
Output: [-1]

Constraints:

  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • 3000 <= preorder[i], inorder[i] <= 3000
  • preorder 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.

Solve

法1

解

前序的第一點就是TreeNode的root

找到中序的root.val 對應值，左邊就是左子樹，右邊就是右子樹

遞迴下去即可

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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]:
        if not preorder or not inorder:
            return None
        root = TreeNode(preorder[0])
        
        n = 0

        for i in range(len(inorder)):
            if inorder[i] == preorder[0]:
                n = i
                break
        root_l_pre = preorder[1:n+1]
        root_r_pre = preorder[n+1:]
        root_l_in = inorder[:n]
        root_r_in = inorder[ n+1 : ]
        
        
        root.left = self.buildTree(root_l_pre,root_l_in)
        root.right = self.buildTree(root_r_pre,root_r_in)

        return root

整理過後

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class Solution:
  def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:

    if not preorder or not inorder:
      return None

    root = TreeNode(preorder[0])
    root_index = inorder.index(preorder[0])

    root.left = self.buildTree(preorder[1:root_index +1], inorder[:root_index ])
    root.right = self.buildTree(preorder[root_index +1:], inorder[root_index +1:])

    return root