Okii honii 3ana problem esmo “ Binary Tree Paths” okiii 

Bahiii honii 3ana problem esmo the “root” of a binary Tree , return all “root-to-leaf” paths 
In any order : 
Q- bahii honii 3ana definition of what is “leaf” ? 

—> a leaf is a node with no children okii 





Bahii honii 3ana “root” = [1,2,3,null,5] 

—> donc honii 3ana problem mt3 enik ttl3 el path mt3 node and just print it okii 

Bahii awl 7aje chnowa howa all the pathis in the Tree 
Path 1 : 1->2->5 // wsolna el leaf node twa okii 
Path  2 : 1 ->3 // wsolan el leaf node okii

Q- donc honi el trick hawaii enikk “traverse” the binary Tree , create the right string format str=”node1 → node2 → node2” okii and push it into the result list okii 

For the “Traversal” part we use dfs okii 
Stack  : okii 
For the string format we just “format “ it okii 
And to store it just append it to the result list okii 

Q- dfs : 

stack  = [(root,””)] 
Result = [] 
While stack:
	node,s=stack.pop()
	current_string=”s”

Q- wait nsef 7je ntsw5er lzemn i check if the node is a leaf (yes or not ) : 

	If is_leaf(node):
		result.append(current_string) 
Q- now let’s try continue the logic for me ? 
	if node.right:
		stack.append(node,”->node.right.val”)
	If node.left:
		stack.append(node,”->node.left.val”) 
Return result 

Q- chof honi awl iteration 3andii anii honii 
Q- mch a7sen w7da wle a5er w7eda hya 
Q- mais n7ebk tkon el pair programmer pattern for me and give me a good feedback okii 
Q- without given me the full slutoon oki

It sounds like you’re off to a solid start with the “Binary Tree Paths” problem, and I can tell you’re following a nice structured plan! You're handling the traversal with DFS and managing the path building with a stack-based approach. Here’s some feedback to consider:

### Feedback

1. **Path String Formatting:** You’re using `current_string = s`, which might not work as expected. Consider if `s` includes the whole path up to the current node, not just the immediate value. If not, think about updating `current_string` to include the `node.val` at each step before pushing to `result`.

2. **DFS Traversal Order:** You’re setting up the right traversal logic, where each time you encounter a node with children, you add the `left` and `right` nodes to the stack along with the path so far. However, double-check your `stack.append` format: when you append `node.right` and `node.left`, the strings might not be fully formatted yet. Ensure they properly concatenate the path so far.

3. **Leaf Node Condition:** Your check for `is_leaf(node)` seems correct, but remember to fully define it as a helper function. In Python, something like `def is_leaf(node): return not node.left and not node.right` could help maintain clarity and avoid redundancy in your code.

4. **Overall Flow:** The logic flow is good, but watch for minor syntax details when adding the right and left children to the stack. Also, consider initializing `current_string` with a `root.val` to start with the root in the path.

Would you like to work through any specific parts together?


Q- bhii honii 3andi mochkma kefm bsh i mantain “ 1 -> 2 -> 5” format here while traversal 
Q- can you give me some hints 

Absolutely! Maintaining the `"1 -> 2 -> 5"` format throughout the traversal can be a bit tricky, but here are a few hints to help keep the path string correctly formatted:

### Hint 1: Update the Path in Each Iteration
Each time you move to a new node, build on the path from the previous node. You can do this by:
   - Adding the current node’s value to the existing path string.
   - If you’re at the root, start with the root’s value.
   - For child nodes, update the path by appending `"->"` followed by the child’s value.

For example, if you’re at node `2` and already have `"1"` in your path, update it to `"1->2"`.

### Hint 2: Avoid Over-Appending "->"
Only add `"->"` if you're moving from a parent node to a child. For a leaf node (with no children), don’t add `"->"` after its value; instead, add it directly to the result.

### Hint 3: Keep Track of the Path with Each Node
Since you’re using DFS, each node in the stack can carry its path string up to that point. So:
   - When you push a child onto the stack, include the updated path string for that child.
   - This will ensure that when you pop a node, you already have the full path for it without having to backtrack.

Would you like a bit more detail on any of these steps?


Q- now i get the full logic honii just i need to start coding okii ? 

// edge case if root is empty 
If not root: 
	Return [] 
// initiate a stack honii 
Stack = [(root,”root.val”)] // push the node and path honii oki  
// create a result list honii oki 
Result = [] 
While stack:
	Node , path=stack.pop()
	current_path+=path
	If is_leaf(node):
		result.append(current_path) 
	// continue filling the stack 
	If node.right:
		stack.append(node.right,”->”+node.right.val) 
	If node.left:
		stack.append(node.left,”->”+node.left.val) 
Return result 


Q- now given the code i create it honii 
Q- did my logic sound yes or not