Binary Trees

• ADT
  • leftChild()
  • rightChild()
  • sibling()
• Properties
  • Level: the set of all nodes at the same depth.
    • The root is the level 0 set
    • The children of the root are the level 1 set.
    • and so on.
    • Some observations: If T is a proper (each node has 0 or 2 children) binary tree, with n nodes, and height of h.
      • The number of external nodes in the tree is at least h+1 andat most 2^h

        Weak induction on the number of nodes in the tree. (h+1 <= e <= 2h)
          Show for some small constant (n=3 nodes) -> Clearly true (e=2  = 1+1 = 21)
          Assume for all n<k
          Prove for n=k.
             Split the tree into two sub trees by removing the root.
             Argue each has height h-1, and therefore h <= e <=  2h-1
             Thus when we recombine the trees, the proposition holds
        

      • The number of internal nodes in T is at least h and at most 2^h-1
      • The total number of nodes in T is at least 2h+1 and at most 2^h+1-1
      • the number of external nodes is one more the the number of internal nodes.
• Traversals: We have already seen these.
• Binary search tree
  • A binary tree T is a binary search tree if foreach node v all elements in the left child of v are less than or equal to the element at v and all elements in the rightchild of v are greater than or equal to the element stored at v.
  • An inorder traversal will print out the items in order.
  • Locating an item in such a tree is O(h), which can be lg(n), or n
  • Inserting new nodes in such a tree is O(h)
• Printing a binary tree.
  • Do an inorder traversal, at each node assign coordinates x and y
  • (y runs from top to bottom, common in graphics)
  • x(v) = number of nodes visited before v
  • y(v) = depth of v.