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Graph and tree search algorithms 

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In computer science, tree traversal (also known as tree search) refers to the process of visiting (examining and/or updating) each node in a tree data structure, exactly once, in a systematic way. Such traversals are classified by the order in which the nodes are visited. The following algorithms are described for a binary tree, but they may be generalized to other trees as well.
Compared to linear data structures like linked lists and onedimensional arrays, which have a canonical method of traversal (namely in linear order), tree structures can be traversed in many different ways. Starting at the root of a binary tree, there are three main steps that can be performed and the order in which they are performed defines the traversal type. These steps (in no particular order) are: performing an action on the current node (referred to as "visiting" the node), traversing to the left child node, and traversing to the right child node.
Traversing a tree involves iterating (looping) over all nodes in some manner. Because from a given node there is more than one possible next node (it is not a linear data structure), then, assuming sequential computation (not parallel), some nodes must be deferred – stored in some way for later visiting. This is often done via a stack (LIFO) or queue (FIFO). As a tree is a selfreferential (recursively defined) data structure, traversal can naturally be described by recursion or, more subtly, corecursion, in which case the deferred nodes are stored implicitly – in the case of recursion, in the call stack.
The name given to a particular style of traversal comes from the order in which nodes are visited. Most simply, does one go down first (depthfirst: first child, then grandchild before second child) or across first (breadthfirst: first child, then second child before grandchildren)? Depthfirst traversal is further classified by position of the root element with regard to the left and right nodes. Imagine that the left and right nodes are constant in space, then the root node could be placed to the left of the left node (preorder), between the left and right node (inorder), or to the right of the right node (postorder). There is no equivalent variation in breadthfirst traversal – given an ordering of children, "breadthfirst" is unambiguous.
For the purpose of illustration, it is assumed that left nodes always have priority over right nodes. This ordering can be reversed as long as the same ordering is assumed for all traversal methods.
Depthfirst traversal is easily implemented via a stack, including recursively (via the call stack), while breadthfirst traversal is easily implemented via a queue, including corecursively.
Beyond these basic traversals, various more complex or hybrid schemes are possible, such as depthlimited searches such as iterative deepening depthfirst search.
There are three types of depthfirst traversal: preorder,^{[1]} inorder,^{[1]} and postorder.^{[1]} For a binary tree, they are defined as operations recursively at each node, starting with the root node as follows:
The trace of a traversal is called a sequentialisation of the tree. No one sequentialisation according to pre, in or postorder describes the underlying tree uniquely. Given a tree with distinct elements, either preorder or postorder paired with inorder is sufficient to describe the tree uniquely. However, preorder with postorder leaves some ambiguity in the tree structure.^{[2]}
To traverse any tree in depthfirst order, perform the following operations recursively at each node:
where n is the number of child nodes. Depending on the problem at hand, the preorder, inorder or postorder operations may be void, or you may only want to visit a specific child node, so these operations are optional. Also, in practice more than one of preorder, inorder and postorder operations may be required. For example, when inserting into a ternary tree, a preorder operation is performed by comparing items. A postorder operation may be needed afterwards to rebalance the tree.
Trees can also be traversed in levelorder, where we visit every node on a level before going to a lower level.
Preorder traversal while duplicating nodes and values can make a complete duplicate of a binary tree. It can also be used to make a prefix expression (Polish notation) from expression trees: traverse the expression tree preorderly.
Inorder traversal is very commonly used on binary search trees because it returns values from the underlying set in order, according to the comparator that set up the binary search tree (hence the name).
Postorder traversal while deleting or freeing nodes and values can delete or free an entire binary tree. It can also generate a postfix representation of a binary tree.
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preorder(node) if node == null then return visit(node) preorder(node.left) preorder(node.right)  iterativePreorder(node) parentStack = empty stack parentStack.push(null) top = node while ( top != null ) visit( top ) if ( top.right != null ) parentStack.push(top.right) if ( top.left != null ) parentStack.push(top.left) top = parentStack.top(); parentStack.pop(); 
inorder(node) if node == null then return inorder(node.left) visit(node) inorder(node.right)  iterativeInorder(node) parentStack = empty stack while (not parentStack.isEmpty() or node ≠ null) if (node ≠ null) parentStack.push(node) node = node.left else node = parentStack.pop() visit(node) node = node.right 
postorder(node) if node == null then return postorder(node.left) postorder(node.right) visit(node)  iterativePostorder(node) parentStack = empty stack lastnodevisited = null while (not parentStack.isEmpty() or node ≠ null) if (node ≠ null) parentStack.push(node) node = node.left else peeknode = parentStack.peek() if (peeknode.right ≠ null and lastnodevisited ≠ peeknode.right) /* if right child exists AND traversing node from left child, move right */ node = peeknode.right else parentStack.pop() visit(peeknode) lastnodevisited = peeknode 
All the above implementations require call stack space proportional to the height of the tree. In a poorly balanced tree, this can be considerable. We can remove the stack requirement by maintaining parent pointers in each node, or by threading the tree (next section).
A binary tree is threaded by making every left child pointer point to the inorder predecessor of the node and every right child pointer point to the inorder successor of the node.
Advantages:
Disadvantages:
Morris traversal is an implementation of inorder traversal that uses threading:
Also, listed below is pseudocode for a simple queue based level order traversal, and will require space proportional to the maximum number of nodes at a given depth. This can be as much as the total number of nodes / 2. A more spaceefficient approach for this type of traversal can be implemented using an iterative deepening depthfirst search.
levelorder(root) q = empty queue q.enqueue(root) while not q.empty do node := q.dequeue() visit(node) if node.left ≠ null then q.enqueue(node.left) if node.right ≠ null then q.enqueue(node.right)
While traversal is usually done for trees with a finite number of nodes (and hence finite depth and finite branching factor) it can also be done for infinite trees. This is of particular interest in functional programming (particularly with lazy evaluation), as infinite data structures can often be easily defined and worked with, though they are not (strictly) evaluated, as this would take infinite time. Some finite trees are too large to represent explicitly, such as the game tree for chess or go, and so it is useful to analyze them as if they were infinite.
A basic requirement for traversal is to visit every node. For infinite trees, simple algorithms often fail this. For example, given a binary tree of infinite depth, a depthfirst traversal will go down one side (by convention the left side) of the tree, never visiting the rest, and indeed if inorder or postorder will never visit any nodes, as it has not reached a leaf (and in fact never will). By contrast, a breadthfirst (levelorder) traversal will traverse a binary tree of infinite depth without problem, and indeed will traverse any tree with bounded branching factor.
On the other hand, given a tree of depth 2, where the root node has infinitely many children, and each of these children has two children, a depthfirst traversal will visit all nodes, as once it exhausts the grandchildren (children of children of one node), it will move on to the next (assuming it is not postorder, in which case it never reaches the root). By contrast, a breadthfirst traversal will never reach the grandchildren, as it seeks to exhaust the children first.
A more sophisticated analysis of running time can be given via infinite ordinal numbers; for example, the breadthfirst traversal of the depth 2 tree above will take ?·2 steps: ? for the first level, and then another ? for the second level.
Thus, simple depthfirst or breadthfirst searches do not traverse every infinite tree, and are not efficient on very large trees. However, hybrid methods can traverse any (countably) infinite tree, essentially via a diagonal argument ("diagonal" – a combination of vertical and horizontal – corresponds to a combination of depth and breadth).
Concretely, given the infinitely branching tree of infinite depth, label the root node the children of the root node the grandchildren and so on. The nodes are thus in a onetoone correspondence with finite (possibly empty) sequences of positive numbers, which are countable and can be placed in order first by sum of entries, and then by lexicographic order within a given sum (only finitely many sequences sum to a given value, so all entries are reached – formally there are a finite number of compositions of a given natural number, specifically 2^{n1}compositions of n = 1;), which gives a traversal. Explicitly:
0: () 1: (1) 2: (1,1) (2) 3: (1,1,1) (1,2) (2,1) (3) 4: (1,1,1,1) (1,1,2) (1,2,1) (1,3) (2,1,1) (2,2) (3,1) (4)
etc.
This can be interpreted as mapping the infinite depth binary tree onto this tree and then applying breadthfirst traversal: replace the "down" edges connecting a parent node to its second and later children with "right" edges from the 1st child to the 2nd child, 2nd child to third child, etc. Thus at each step one can either go down (append a (,1) to the end) or go right (add 1 to the last number) (except the root, which is extra and can only go down), which shows the correspondence between the infinite binary tree and the above numbering; the sum of the entries (minus 1) corresponds to the distance from the root, which agrees with the 2^{n1} nodes at depth n1 in the infinite binary tree (2 corresponds to binary).