Traversing a Three-Node Tree

Let’s look at a simple example to get an idea of how this recursive traversal routine works. Imagine traversing a tree with only three nodes: a root (A), with a left child

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inOrder (A)

1. Call inOrder (B) 2. Visit A 3. Call inOrder (C)

inOrder (B)

inOrder (C)

1. Call inOrder (null)

2. Visit B

2. Visit C

3. Call inOrder (null)

inOrder (null)

inOrder (null) Returns

inOrder (null)

FIGURE 8.9 The inOrder() method applied to a three-node tree. We start by calling inOrder() with the root A as an argument. This incarnation of

inOrder() we’ll call inOrder(A) . inOrder(A) first calls inOrder() with its left child, B, as an argument. This second incarnation of inOrder() we’ll call inOrder(B) .

inOrder(B) now calls itself with its left child as an argument. However, it has no left child, so this argument is null . This creates an invocation of inorder() we could call inOrder(null) . There are now three instances of inOrder() in existence: inOrder(A) , inOrder(B) , and inOrder(null) . However, inOrder(null) returns immediately when it finds its argument is null . (We all have days like that.)

Now inOrder(B) goes on to visit B; we’ll assume this means to display it. Then inOrder(B) calls inOrder() again, with its right child as an argument. Again this argument is null , so the second inorder(null) returns immediately. Now inOrder(B) has carried out steps 1, 2, and 3, so it returns (and thereby ceases to exist).

384 CHAPTER 8 Binary Trees

Now we’re back to inOrder(A) , just returning from traversing A’s left child. We visit A and then call inOrder() again with C as an argument, creating inOrder(C) . Like inOrder(B) , inOrder(C) has no children, so step 1 returns with no action, step 2 visits

C, and step 3 returns with no action. inOrder(B) now returns to inOrder(A) . However, inOrder(A) is now done, so it returns and the entire traversal is complete.

The order in which the nodes were visited is A, B, C; they have been visited inorder. In a binary search tree this would be the order of ascending keys.

More complex trees are handled similarly. The inOrder() function calls itself for each node, until it has worked its way through the entire tree.

Traversing with the Workshop Applet

To see what a traversal looks like with the Workshop applet, repeatedly press the Trav button. (You don’t need to type in any numbers.)

Here’s what happens when you use the Tree Workshop applet to traverse inorder the tree shown in Figure 8.10. This is slightly more complex than the three-node tree seen previously. The red arrow starts at the root. Table 8.1 shows the sequence of node keys and the corresponding messages. The key sequence is displayed at the bottom of the Workshop applet screen.

FIGURE 8.10 Traversing a tree inorder.

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TABLE 8.1 Workshop Applet Traversal

Red List of Step

Arrow Nodes Number

on Node

Will check left child 2 30 Will check left child 3 20 Will check left child 4 20 Will visit this node 5 20 Will check right child

6 20 Will go to root of previous subtree

7 30 Will visit this node

8 30 Will check right child

9 40 Will check left child

10 40 Will visit this node

11 40 Will check right child

12 40 Will go to root of previous subtree

13 50 Will visit this node

14 50 Will check right child

15 60 Will check left child

16 60 Will visit this node

17 60 Will check right child

18 60 Will go to root of previous subtree

19 50 Done traversal

It may not be obvious, but for each node, the routine traverses the node’s left subtree, visits the node, and traverses the right subtree. For example, for node 30 this happens in steps 2, 7, and 8.

The traversal algorithm isn’t as complicated as it looks. The best way to get a feel for what’s happening is to traverse a variety of different trees with the Workshop applet.

Traversing a Three-Node Tree