In recursive DFS traversal, we have three basic elements to traverse: root node, left subtree, and right subtree. Each traversal process nodes in a different order, where recursive code is simple and easy to visualize i.e. one function parameter and 3-4 lines of code. The critical question is: How can we convert the code of recursive tree traversal into an iterative tree traversal? Let’s think.
In recursive code, the compiler uses the call stack in the background to convert it into iterative code. This call stack stores information like local variables, input parameters, the state of function calls, etc. So, for iterative implementation, we need to mimic what the compiler does when we run recursive code! In other words: We need to use an explicit stack to simulate the execution of recursion.
Sometimes, the iterative implementation using a stack becomes complex due to many input variables, additional operations, and the complex nature of recursive calls. In some situations like binary tree traversal, recursion is simple, and we can easily convert it into iterative code.
Preorder and inorder traversals are tail-recursive, i.e., there are no extra operations after the final recursive call. So the implementation of preorder and inorder traversals using a stack is simple and easy to understand.
On the other side, postorder traversal is non-tail recursive because there is one extra operation after the last recursive call: We process the root node. So implementing postorder using a stack can be a little tricky. But if we follow the order of nodes in postorder, it can be easy to visualize.
One idea is clear: To simulate the recursive tree traversal into an iterative traversal, we need to understand the flow of recursive calls. In recursive tree traversal, we visit each node three times:
Using the above insight, let's simulate the iterative implementation of each tree traversal using stack. We will be using the following node structure of the binary tree.
//Node structure in C++
class TreeNode
{
public:
int data;
TreeNode* left;
TreeNode* right;
TreeNode(int value)
{
data = value;
left = NULL;
right = NULL;
}
};
//Node structure in Java
class TreeNode
{
int data;
TreeNode left;
TreeNode right;
public TreeNode(int value)
{
data = value;
left = null;
right = null;
}
}
//Node structure in Python
class TreeNode:
def __init__(self, value):
self.data = value
self.left = None
self.right = None
Let's revise the preorder traversal: We first process the root node, then traverse the left subtree, and finally, traverse the right subtree. So tracking the correct flow of recursion will give us a better picture.
The overall idea of iterative preorder traversal using a stack: When we visit any node for the first time, we process it, push that node into the stack (to process the right subtree of that node later), and go to the left child. If there is no left child, we remove the top node from the stack and go to the right child of that node. Now we continue the same process for the subtree rooted at the right child of the popped node.
Condition 1: If (currNode != NULL)
We process the currNode and push it into treeStack. Now we move to the left child i.e. currNode = currNode->left. This process will continue in a loop until we reach the leftmost end of the binary tree i.e. currNode == NULL. At this stage, all ancestors of that node will be available on treeStack.
If currNode != NULL:
Condition 2: If (currNode == NULL)
Now we have reached the leftmost end of the binary tree, so we move to the parent by removing the top node from treeStack. At this stage, we need to traverse the right subtree of the top node. So we assign currNode to the right child of the removed top node and continue the above process.
If currNode == NULL:
void iterativePreorder(TreeNode* root)
{
if (root == NULL)
return;
stack<TreeNode*> treeStack;
TreeNode* currNode = root;
TreeNode* prevNode = NULL;
while (treeStack.empty() == false || currNode != NULL)
{
if (currNode != NULL)
{
cout << currNode->data << " ";
treeStack.push(currNode);
currNode = currNode->left;
}
else
{
prevNode = treeStack.top();
treeStack.pop();
currNode = prevNode->right;
}
}
}
def iterativePreorder(root):
if root is None:
return
treeStack = []
currNode = root
while treeStack or currNode:
if currNode:
print(currNode.data, end=" ")
treeStack.append(currNode)
currNode = currNode.left
else:
prevNode = treeStack.pop()
currNode = prevNode.right
public void iterativePreorder(TreeNode root)
{
if (root == null)
return;
Stack<TreeNode> treeStack = new Stack<>();
TreeNode currNode = root;
while (!treeStack.empty() || currNode != null)
{
if (currNode != null)
{
System.out.print(currNode.data + " ");
treeStack.push(currNode);
currNode = currNode.left;
}
else
{
TreeNode prevNode = treeStack.pop();
currNode = prevNode.right;
}
}
}
In the above code, we are storing any node in the stack to access the right subtree of that node later in preorder. So instead of storing that node, we can directly store the right child of that node. This will help us to optimize the above solution by reducing the stack space.
void iterativePreorder(TreeNode *root)
{
if (root == NULL)
return;
stack<TreeNode *> treeStack;
TreeNode *currNode = root;
while (treeStack.empty() == false || currNode != NULL)
{
if (currNode)
{
cout << currNode->data << " ";
if (currNode->right != NULL)
treeStack.push(currNode->right);
currNode = currNode->left;
}
else
{
currNode = treeStack.top();
treeStack.pop();
}
}
}
def iterativePreorder(root):
if root is None:
return
treeStack = []
currNode = root
while treeStack or currNode:
if currNode:
print(currNode.data, end=" ")
if currNode.right is not None:
treeStack.append(currNode.right)
currNode = currNode.left
else:
currNode = treeStack.pop()
public void iterativePreorder(TreeNode root)
{
if (root == null)
return;
Stack<TreeNode> treeStack = new Stack<>();
TreeNode currNode = root;
while (!treeStack.empty() || currNode != null)
{
if (currNode != null)
{
System.out.print(currNode.data + " ");
if (currNode.right != null)
treeStack.push(currNode.right);
currNode = currNode.left;
}
else
currNode = treeStack.pop();
}
}
If we simplify the preorder traversal for each node: We process the node first and then process the left and right child. To track this order using a stack, we can push the right child before the left child to ensure that the left subtree is processed first. In other words, if we pop a node from the stack, the left child comes before the right child.
void iterativePreorder(TreeNode *root)
{
if (root == NULL)
return;
stack<TreeNode *> treeStack;
TreeNode *currNode;
treeStack.push(root);
while (treeStack.empty() == false)
{
currNode = treeStack.top();
treeStack.pop();
cout << currNode->data << " ";
if (currNode->right != NULL)
treeStack.push(currNode->right);
if (currNode->left != NULL)
treeStack.push(currNode->left);
}
}
Each node is pushed into and popped out of the stack only once, so we are doing a constant number of operations for each node. The time complexity is n* O(1) = O(n). We are using one stack, and the size of the stack depends on the height of the binary tree. So, the space complexity is O(h).
In an inorder traversal, we first process the left subtree, then the root node, and finally the right subtree. How can we simulate this process using a stack? Let’s understand this by exploring the flow of recursion.
Here is an idea for iterative simulation using a stack: Before processing the left subtree of any node, we need to save that node on the stack (to process that node and the right subtree of that node later). Then we go to the left child of that node. After processing all nodes in the left subtree, we pop the node from the top of the stack, process it, and go to the right child of that node to traverse the right subtree. Let's understand it via clear implementation steps.
Condition 1: If (currNode != NULL)
We push currNode onto the stack to process currNode and the right subtree later. Next, we move to the left child by assigning currNode = currNode->left and push it onto the stack. This step continues in a loop until we reach the leftmost end of the binary tree, i.e. currNode == NULL.
if (currNode != NULL)
{
treeStack.push(currNode)
currNode = currNode->left
}
Condition 2: If (currNode == NULL)
Now we have reached the leftmost end of the binary tree, so we move to the parent of that node by popping a node from the top of treeStack. We assign it to currNode and process it. At this stage, we need to traverse the right subtree of the parent node, so we assign currNode with the right child of the popped node and continue the above process in a loop. Think!
if (currNode == NULL)
{
currNode = treeStack.pop()
process(currNode->data)
currNode = currNode->right
}
void iterativeInorder(TreeNode* root)
{
if (root == NULL)
return;
stack<TreeNode*> treeStack;
TreeNode* currNode = root;
while (treeStack.empty() == false || currNode != NULL)
{
if (currNode != NULL)
{
treeStack.push(currNode);
currNode = currNode->left;
}
else
{
currNode = treeStack.top();
treeStack.pop();
cout << currNode->val << " ";
currNode = currNode->right;
}
}
}
Each node is pushed into and popped out of the treeStack only once, so we are doing a constant number of operations for each node. The time complexity is n * O(1) = O(n).
The space complexity is O(h) since we are using one stack, and the size of the stack depends on the height of the binary tree.
In a postorder traversal, we first process the left subtree, then the right subtree, and finally the root node. In other words, we process all the nodes in the left and right subtrees before processing any node. How do we simulate this process using a stack? Let's track the flow of recursion.
Here is an idea of iterative simulation using stack: Before processing the left subtree of any node, we need to save two things on the stack: (1) the right child of the current node to process the right subtree after the traversal of the left subtree, and (2) the current node, so that we can process it after the traversal of the right subtree. To simulate this, we can use two separate stacks to store these two nodes. But how do we track the nodes in a postorder fashion? Let’s understand it clearly via the implementation.
Condition 1: If (currNode != NULL)
If the right child of the currNode is not NULL, then we push the right child into the rightChildStack. Now we push the currNode into the mainStack and go to the left child. We continue the same process until we reach the leftmost end of the tree, i.e., currNode == NULL.
if(currNode != NULL)
{
if(currNode->right != NULL)
rightChildStack.push(currNode->right)
mainStack.push(currNode)
currNode = currNode->left
}
Condition 2: If (currNode == NULL)
Now, we access the parent node of currNode, which is at the top of the mainStack. We assign the top node to currNode, but before processing it, we need to first process the nodes in the right subtree.
currNode = mainStack.top()
if(!rightChildStack.isEmpty() && currNode->right == rightChildStack.top())
currNode = rightChildStack.pop()
else
{
process(currNode->data)
mainStack.pop()
currNode = NULL
}
void iterativePostorder(TreeNode* root)
{
if (root == NULL)
return;
stack<TreeNode*> mainStack;
stack<TreeNode*> rightChildStack;
TreeNode* currNode = root;
while (mainStack.empty() == false || currNode != NULL)
{
if (currNode != NULL)
{
if (currNode->right != NULL)
rightChildStack.push(currNode->right);
mainStack.push(currNode);
currNode = currNode->left;
}
else
{
currNode = mainStack.top();
if (rightChildStack.empty() == false && currNode->right == rightChildStack.top())
{
currNode = rightChildStack.pop();
}
else
{
cout << currNode->val << " ";
mainStack.pop();
currNode = NULL;
}
}
}
}
Each node is pushed into and popped out of the mainStack only once. Similarly, in the worst case, each node is pushed into the rightChildStack at most once. So we are doing a constant number of operations for each node. The time complexity is n * O(1) = O(n).
Space complexity: We are using two different stacks, where the size of each stack depends on the height of the binary tree. Think! So the space complexity = O(h).
Can we optimize the above idea of postorder traversal and do it using a single stack? Think! In the code below, we use a single stack to perform iterative postorder traversal and keep track of the previously visited node using the prevNode variable.
void postorderTraversalSingleStack(TreeNode* root)
{
if (root == NULL)
return;
stack<TreeNode*> treeStack;
TreeNode* prevNode = NULL;
while (root != NULL || !treeStack.empty())
{
if (root != NULL)
{
treeStack.push(root);
root = root->left;
}
else
{
TreeNode* currNode = treeStack.top();
if (currNode->right != NULL && currNode->right != prevNode)
{
// go to right child if it exists and has not been visited
root = currNode->right;
}
else
{
// visit the current node
cout << currNode->val << " ";
prevNode = currNode;
treeStack.pop();
}
}
}
}
void iterativePostorderUsingSet(TreeNode* root)
{
if (root == NULL)
return;
stack<TreeNode*> treeStack;
unordered_set<TreeNode*> visitedNodeSet;
TreeNode* currNode = root;
while (!treeStack.empty() || currNode != NULL)
{
if (currNode != NULL)
{
if (visitedNodeSet.count(currNode) > 0)
{
// currNode has already been visited
cout << currNode->val << " ";
currNode = NULL;
}
else
{
// first visit to currNode
treeStack.push(currNode);
if (currNode->right != NULL)
treeStack.push(currNode->right);
visitedNodeSet.insert(currNode);
currNode = currNode->left;
}
}
else
{
// currNode is NULL, pop next node from stack
currNode = treeStack.top();
treeStack.pop();
}
}
}
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