Tree traversal

1
Tree traversal

• void TraverseTree(BintreePtr Root)
• if(Root ! NULL)
• TraverseTree(Root.getLeft())
• System.out.print(Root.getValue()"\t")
• TraverseTree(Root.getRight())

Base case if the tree is empty do nothing
2
Tree traversal

• void TraverseTree(BintreePtr Root)
• if(Root ! NULL)
• TraverseTree(Root.getLeft())
• System.out.print(Root.getValue()"\t")
• TraverseTree(Root.getRight())

Recursively call TraverseTree() to traverse left
subtree...
3
Tree traversal

• void TraverseTree(BintreePtr Root)
• if(Root ! NULL)
• TraverseTree(Root.getLeft())
• System.out.print(Root.getValue()"\t")
• TraverseTree(Root.getRight())

Print current node (can alter this step to
do anything to current node)
4
Tree traversal

• void TraverseTree(BintreePtr Root)
• if(Root ! NULL)
• TraverseTree(Root.getLeft())
• System.out.print(Root.getValue()"\t")
• TraverseTree(Root.getRight())

Use recursion to traverse right subtree
5
Other types of traversal

• Inorder traversal (Left-Node-Right LNR)
• Preorder traversal (Node-Left-Right NLR)
• Postorder traversal (Left-Right-Node LRN)

6
Inorder traversal (LNR)
3
1
9
12
2
Left
10
Right
7
Inorder traversal (LNR)
3
1
9
12
2
Left 1 2
10
Right 9 10 12
Traversal 1 2 3 9 10 12
8
Preorder traversal (NLR)
3
1
9
12
2
Left
10
Right
9
Preorder traversal (NLR)
3
1
9
12
2
Left 1 2
10
Right 9 12 10
Traversal 3 1 2 9 12 10
10
Postorder traversal (LRN)
3
1
9
12
2
Left
10
Right
11
Postorder traversal (LRN)
3
1
9
12
2
Left 2 1
10
Right 10 12 9
Traversal 2 1 10 12 9 3
12
Deleting nodes from a tree

• Ensure that BST properties still hold after
  deletion
• Three types of nodes to delete
• Leaf
• Node with one child
• Node with 2 children

13
Deleting a leaf
X
Delete D
V
F
C
T
Z
U
Y
W
E
D
J
Q
14
Deleting a leaf
X
Delete D
V
F
C
T
Z
U
Y
W
E
J
Q
15
Deleting a node with 1 child
X
Delete F Connect child to parent of F
V
F
C
T
Z
U
Y
W
E
J
Q
16
Deleting a node with 1 child
X
Delete F
V
Z
C
T
Y
W
U
E
J
Q
17
Deleting a node with 2 children
X
Delete X
V
Z
C
T
Y
W
U
E
J
Q
Two ways of doing it...go left or go right
18
Go left...
X
Delete X
V
Z
C
T
Y
W
U
E
J
Q

• Find RIGHTMOST node in LEFT subtree
• Usually referred to as InOrder Predecessor
• Has to have AT MOST one child

19
Go left...
T
Delete X
V
Z
C
T
Y
W
U
E
J
Q

• Copy VALUE in this node to replace value in
  delete node

20
Go left...
T
Delete X
V
Z
C
T
Y
W
U
E
J
Q

• Copy VALUE in this node to replace value in
  delete node
• Then delete InOrder Predecessor

21
Go left...
T
Delete X
V
Z
U
C
Y
W
E
J
Q
Deletion completed
22
Go right...
X
Delete X
V
Z
C
T
Y
W
U
E
J
Q

• Find LEFTMOST node in RIGHT subtree
• Usually referred to as InOrder Successor
• Has to have AT MOST one child

23
Go right...
E
Delete X
V
Z
C
T
Y
W
U
E
J
Q

• Copy VALUE in this node to replace value in
  delete node

24
Go right...
E
Delete X
V
Z
C
T
Y
W
U
E
J
Q

• Copy VALUE in this node to replace value in
  delete node
• Then delete InOrder Successor

25
Go right...
E
Delete X
V
Z
C
T
Y
W
U
J
Q
Deletion completed
26
Counting nodes in a tree

• int CountNodes(BintreePtr Root)
• if(Root ! NULL)
• return 1 CountNodes(Root.getLeft())
• CountNodes(Root.getRight())
• else
• return 0

27
Counting nodes in a tree

• int CountNodes(BintreePtr Root)
• if(Root ! NULL)
• return 1 CountNodes(Root.getLeft())
• CountNodes(Root.getRight())
• else
• return 0

Base case Check that Root isn't NULL
28
Counting nodes in a tree

• int CountNodes(BintreePtr Root)
• if(Root ! NULL)
• return 1 CountNodes(Root.getLeft())
• CountNodes(Root.getRight())
• else
• return 0

Number of nodes is 1 (for current node)...
29
Counting nodes in a tree

• int CouNpNodes(BintreePtr Root)
• if(Root ! NULL)
• return 1 CountNodes(Root.getLeft())
• CountNodes(Root.getRight())
• else
• return 0

Number of nodes in the left subtree...
30
Counting nodes in a tree

• int CountNodes(BintreePtr Root)
• if(Root ! NULL)
• return 1 CountNodes(Root.getLeft())
• CountNodes(Root.getRight())
• else
• return 0

Number of nodes in the right subtree.
31
Counting nodes in a tree

• int CountNodes(BintreePtr Root)
• if(Root ! NULL)
• return 1 CountNodes(Root.getLeft())
• CountNodes(Root.getRight())
• else
• return 0

If Root is NULL, there are no nodes, so return 0
32
Finding the depth of a tree

• The depth of a tree is the maximum path length
  from the root to a leaf
• Depth of tree 5

33
Finding the depth of a tree

• Recursive algorithm
• if(Root ! NULL) / base case /
• Find the depth of the left branch
• ...and of right branch
• Depth 1 max of depth(left) and
  depth(right)
• Else
• Depth 0

34
Finding the depth of a tree
If Root is NULL, there are no nodes, so return 0

• int Depth(BintreePtr Root)
• int DepthLeft, DepthRight
• if(Root ! NULL)
• DepthLeft Depth(Root.getLeft())
• DepthRight Depth(Root.getRight())
• if (DepthLeft gt DepthRight)
• return 1 DepthLeft
• else
• return 1 DepthRight
• else
• return 0

35
Finding the depth of a tree
Recursively find depths of left and right branches

• int Depth(BintreePtr Root)
• int DepthLeft, DepthRight
• if(Root ! NULL)
• DepthLeft Depth(Root.getLeft())
• DepthRight Depth(Root.getRight())
• if (DepthLeft gt DepthRight)
• return 1 DepthLeft
• else
• return 1 DepthRight
• else
• return 0

36
Finding the depth of a tree
Find which branch has larger depth

• int Depth(BintreePtr Root)
• int DepthLeft, DepthRight
• if(Root ! NULL)
• DepthLeft Depth(Root.getLeft())
• DepthRight Depth(Root.getRight())
• if (DepthLeft gt DepthRight)
• return 1 DepthLeft
• else
• return 1 DepthRight
• else
• return 0

37
Finding the depth of a tree
...and return 1 (for current node) larger depth

• int Depth(BintreePtr Root)
• int DepthLeft, DepthRight
• if(Root ! NULL)
• DepthLeft Depth(Root.getLeft())
• DepthRight Depth(Root.getRight())
• if (DepthLeft gt DepthRight)
• return 1 DepthLeft
• else
• return 1 DepthRight
• else
• return 0

38
Finding the depth of a tree
If tree is empty, depth is zero

• int Depth(BintreePtr Root)
• int DepthLeft, DepthRight
• if(Root ! NULL)
• DepthLeft Depth(Root.getLeft())
• DepthRight Depth(Root.getRight())
• if (DepthLeft gt DepthRight)
• return 1 DepthLeft
• else
• return 1 DepthRight
• else
• return 0

39
Counting the leaves in a tree

• Recursive algorithm
• if(Root ! NULL) / base case /
• If (Root is a leaf)
• return 1
• Else
• return Leaves(Root.getLeft()) Leaves(Root.getRig
  ht())
• Else
• return 0

40
Counting the leaves in a tree

• int Leaves(BintreePtr Root)
• if(Root ! NULL)
• if (Root.getLeft() NULL Root.getRight()
  NULL)
• return 1
• else
• return Leaves(Root.getLeft())
• Leaves(Root.getRight())
• else
• return 0

41
Saving a tree to a file

• Other data structures (stacks, queues, lists,
  etc) easy to save to disk
• linear data structures
• traverse list or array, save to file in order
• Trees are 2D data structures
• File structure is linear
• What to do?

42
Saving a tree to a file

• Traverse the tree
• Save data in traversal order
• Which traversal order?
• Inorder results in right-degenerate tree
• PostOrder not right-degenerate but not the same
  as the tree we saved
• PreOrder ?
• To load from a file
• Read each item from file
• Insert into (initially empty) tree

43
Save tree
3
1
9
12
2
10
Preorder traversal gives 3 1 2 9 12 10
44
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Output -gt E V Z C T Y W U Q J
45
Breath-first Traversal

• BreathFirst(Node Root)
• Queue Q new Queue()
• if(Root ! NULL) Q.Join(Root) // insert
  node into Q
• while ( !Q.isempty())
• Root Q.Leave() // remove next node
• Visit(Root) // visit node
• // insert left subtree in Q
• if(Root.getLeft() ! NULL) Q.Join(Root.getLeft
  ())
• // insert right subtree in Q
• if(Root.getRight() ! NULL) Q.Join(Root.getRig
  ht())

46
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt E
updated Q -gt V Z
Output -gt E
47
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt V Z
updated Q -gt Z C T
Output -gt E V
48
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt Z C T
updated Q -gt C T Y W
Output -gt E V Z
49
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt C T Y W
updated Q -gt T Y W
Output -gt E V Z C
50
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt T Y W
updated Q -gt Y W U
Output -gt E V Z C T
51
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt Y W U
updated Q -gt W U
Output -gt E V Z C T Y
52
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt W U
updated Q -gt U Q J
Output -gt E V Z C T Y W
53
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt U Q J
updated Q -gt Q J
Output -gt E V Z C T Y W U
54
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt Q J
updated Q -gt J
Output -gt E V Z C T Y W U Q
55
Breath-first Traversal
E
V
Z
C
T
Y
W
U
J
Q
Q -gt J
updated Q -gt empty
Output -gt E V Z C T Y W U Q J
56
Reconstructing a tree from its traversals

• Can draw a binary tree given
• Inorder traversal (essential)
• One of
• Preorder traversal
• Postorder traversal

57
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Start with postorder
• Last item (X) must be root of tree

X
58
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Locate X in inorder traversal
• Items to left of X are in left subtree
• Items to right of X are in right subtree

X
59
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Construct left subtree
• Find root of left subtree from postorder (V)

X
V
60
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find V in inorder traversal
• C is left child of V

X
V
C
61
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• U T are right subtree of V
• Use postorder to find root of right subtree (T)

X
V
C
T
62
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find T in inorder
• U is left child of T

X
V
C
T
U
63
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Now consider right subtree of X
• Find root from postorder (F)

X
V
F
C
T
U
64
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find F in inorder
• All other nodes are in right subtree of F

X
V
F
C
T
U
65
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find right subtree of F in postorder
• Root is Z

X
V
F
C
T
Z
U
66
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find Z in inorder
• E Y D in left subtree of Z
• Q W J in right subtree of Z

X
V
F
C
T
Z
U
67
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Consider left subtree ( E Y D)
• Find root from postorder (Y)

X
V
F
C
T
Z
U
Y
68
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find Y in inorder
• E is left child of Y
• D is right child of Y

X
V
F
C
T
Z
U
Y
E
D
69
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Consider right subtree of Z
• Find root from postorder (W)

X
V
F
C
T
Z
U
Y
W
E
D
70
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Find W in inorder
• Q is left child of W
• J is right child of W

X
V
F
C
T
Z
U
Y
W
E
D
J
Q
71
Inorder C V U T X F E Y D Z Q W JPostorder C U
T V E D Y Q J W Z F X

• Tree complete

X
V
F
C
T
Z
U
Y
W
E
D
J
Q