105. Construct Binary Tree from Preorder and Inorder Traversal

# Time:  O(n)
# Space: O(n)

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    # @param preorder, a list of integers
    # @param inorder, a list of integers
    # @return a tree node
    def buildTree(self, preorder, inorder):
        lookup = {}
        for i, num in enumerate(inorder):
            lookup[num] = i
        return self.buildTreeRecu(lookup, preorder, inorder, 0, 0, len(inorder))

    def buildTreeRecu(self, lookup, preorder, inorder, pre_start, in_start, in_end):
        if in_start == in_end:
            return None
        node = TreeNode(preorder[pre_start])
        i = lookup[preorder[pre_start]]
        node.left = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1, in_start, i)
        node.right = self.buildTreeRecu(lookup, preorder, inorder, pre_start + 1 + i - in_start, i + 1, in_end)
        return node


# time: O(n)
# space: O(n)
class Solution2(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        preorder_iterator = iter(preorder)
        inorder_lookup = {n: i for i, n in enumerate(inorder)}
        
        def helper(start, end):
            if start > end:
                return None
            
            root_val = next(preorder_iterator)
            root = TreeNode(root_val)
            idx = inorder_lookup[root_val]
            root.left = helper(start, idx-1)
            root.right = helper(idx+1, end)
            return root

        return helper(0, len(inorder)-1)

Solution from kamyu104/LeetCode-Solutions · MIT