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Problem Solution First element of preorder will always be the root of the tree. Inorder traversal searches left tree and then right tree, so we can find the root element in the list and divide the list up into left and right side(trees). Now we can build those trees. Repeat this process with the root of left tree, which is the second element in the preorder list. Then use the third element as th..

Problem Recursive Solution Time Complexity O(N) Space Complexity O(N) worst case, O(logN) average case (recursive space of tree height) # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): def inorderTraversal(self, root): """ :type root: TreeNode :rtype:..