889. Construct Binary Tree from Preorder and Postorder Traversal

Problem Link

Description

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Given two integer arrays, preorder and postorder where preorder is the preorder traversal of a binary tree of distinct values and postorder is the postorder traversal of the same tree, reconstruct and return the binary tree.

If there exist multiple answers, you can return any of them.

 

Example 1:

Input: preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]

Example 2:

Input: preorder = [1], postorder = [1]
Output: [1]

 

Constraints:

• 1 <= preorder.length <= 30
• 1 <= preorder[i] <= preorder.length
• All the values of preorder are unique.
• postorder.length == preorder.length
• 1 <= postorder[i] <= postorder.length
• All the values of postorder are unique.
• It is guaranteed that preorder and postorder are the preorder traversal and postorder traversal of the same binary tree.

Solution

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Python3

# 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 constructFromPrePost(self, preorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        N = len(preorder)
        mp = {}
 
        for i, x in enumerate(postorder):
            mp[x] = i
 
        # pre = [root, left, right]
        # post = [left, right, root]
        
        def go(preLeft, preRight, postLeft, postRight):
            if preLeft > preRight: return None
 
            node = TreeNode(preorder[preLeft])
 
            if preLeft == preRight:
                return node
 
            val = preorder[preLeft + 1]
            idx = mp[val]
            offset = idx - postLeft
 
            node.left = go(preLeft + 1, preLeft + 1 + offset, postLeft, idx)
            node.right = go(preLeft + 1 + offset + 1, preRight, idx + 1, postRight)
 
            return node
        
        return go(0, N - 1, 0, N - 1)