Binary Trees. 2 Linear data structures Here are some of the data structures we have studied so far: –Arrays –Singly-linked lists and doubly-linked lists.

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Presentation transcript:

Binary Trees

2 Linear data structures Here are some of the data structures we have studied so far: –Arrays –Singly-linked lists and doubly-linked lists –Stacks, queues, and deques –Sets These all have the property that their elements can be adequately displayed in a straight line Binary trees are one of the simplest nonlinear data structures

3 Parts of a binary tree A binary tree is composed of zero or more nodes Each node contains: –A value (some sort of data item) –A reference or pointer to a left child (may be null ), and –A reference or pointer to a right child (may be null ) A binary tree may be empty (contain no nodes) If not empty, a binary tree has a root node –Every node in the binary tree is reachable from the root node by a unique path A node with neither a left child nor a right child is called a leaf –In some binary trees, only the leaves contain a value

4 Picture of a binary tree a bc d e ghi l f jk

5 Size and depth The size of a binary tree is the number of nodes in it –This tree has size 12 The depth of a node is its distance from the root –a is at depth zero –e is at depth 2 The depth of a binary tree is the depth of its deepest node –This tree has depth 4 a bc def ghijk l

6 Balance A binary tree is balanced if every level above the lowest is “full” (contains 2 n nodes) In most applications, a reasonably balanced binary tree is desirable a bc defg hij A balanced binary tree a b c d e f gh ij An unbalanced binary tree

7 Binary search in an array Look at array location (lo + hi)/ Searching for 5: (0+6)/2 = 3 hi = 2; (0 + 2)/2 = 1 lo = 2; (2+2)/2= Using a binary search tree

8 Tree traversals A binary tree is defined recursively: it consists of a root, a left subtree, and a right subtree To traverse (or walk) the binary tree is to visit each node in the binary tree exactly once Tree traversals are naturally recursive Since a binary tree has three “parts,” there are six possible ways to traverse the binary tree: –root, left, right –left, root, right –left, right, root –root, right, left –right, root, left –right, left, root

9 Preorder traversal In preorder, the root is visited first Here’s a preorder traversal to print out all the elements in the binary tree: public void preorderPrint(BinaryTree bt) { if (bt == null) return; System.out.println(bt.value); preorderPrint(bt.leftChild); preorderPrint(bt.rightChild); }

10 Inorder traversal In inorder, the root is visited in the middle Here’s an inorder traversal to print out all the elements in the binary tree: public void inorderPrint(BinaryTree bt) { if (bt == null) return; inorderPrint(bt.leftChild); System.out.println(bt.value); inorderPrint(bt.rightChild); }

11 Postorder traversal In postorder, the root is visited last Here’s a postorder traversal to print out all the elements in the binary tree: public void postorderPrint(BinaryTree bt) { if (bt == null) return; postorderPrint(bt.leftChild); postorderPrint(bt.rightChild); System.out.println(bt.value); }

12 Tree traversals using “flags” The order in which the nodes are visited during a tree traversal can be easily determined by imagining there is a “flag” attached to each node, as follows: To traverse the tree, collect the flags: preorderinorderpostorder A BC DEFG A BC DEFG A BC DEFG A B D E C F G D B E A F C G D E B F G C A

13 Copying a binary tree In postorder, the root is visited last Here’s a postorder traversal to make a complete copy of a given binary tree: public BinaryTree copyTree(BinaryTree bt) { if (bt == null) return null; BinaryTree left = copyTree(bt.leftChild); BinaryTree right = copyTree(bt.rightChild); return new BinaryTree(bt.value, left, right); }

14 Other traversals The other traversals are the reverse of these three standard ones –That is, the right subtree is traversed before the left subtree is traversed Reverse preorder: root, right subtree, left subtree Reverse inorder: right subtree, root, left subtree Reverse postorder: right subtree, left subtree, root

15 The End