- Stack: push curr, push right, push left. Given a binary tree, return the preorder traversal of its nodes' values. BFS. Want the sequence in root, left, and right. Queue? NO. After root.left is ...
You may assume that duplicates do not exist in the tree. See 'Construct tree from inorder + postorder' as example. This problem uses divide and conquer idea as well. For preorder: the front node is ...
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