Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Context that can be derived from the question?

1. Longest / farthest leaf nodes can be found using depth first search method.
2. There are 3 primary way to solve DFS:
3. PreOrder
4. InOrder (Go down 1 direction until it hits null, then come back and grab it's direct parent
5. PostOrder (Go down both direction and only when both direction are null grab it's parent)
6. Le't use one of these

PreOrder  (Grab as you)
Tree:
    1
   / \
  2   3
 / \
4   5


PreOrder: [1, 2, 4, 5, 3]


InOrder
(Go down 1 direction until it hits null, then come back and grab it's direct parent)

Tree:
    1
   / \
  2   3
 / \
4   5
InOrder: [4, 2, 5, 1, 3]




PostOrder:
Go down both direction and only when both direction are null grab it's parent
Tree:
    1
   / \
  2   3
 / \
4   5

PostOrder: [4, 5, 2, 3, 1]

We know that to calculate the max depth of binary tree we need to find out the both left and right sub tree from root.

So it naturally aligns with PostOrder DFS.

The question is what will help us only traverse or account the the longest distance only?

function postOrder(node, result){
    if(node.left){
        postOrder(node.left, result)
    }
    if(node.right){
        postOrder(node.right, result)
    }
    result.push(node.value)
    return result
}

For us to calculate:

The maximum depth of a binary tree is the length of the longest path from the root node to any leaf node.

1. Break Down the Problem

 PostOrder traversal:

1. Left subtree depth is calculated first.
2. Right subtree depth is calculated next.
3. Current node depth is calculated as 1 + max(leftDepth, rightDepth).
4. Also shift focus on depth calculation:
5. If a node is null, return 0 (base case for leaf nodes).

Thus we can conclude:

function postOrder(node, length){
    if(!node){
        return 0
    }
    let leftDepth = postOrder(node.left)
    let rightDepth =  postOrder(node.right)
    return 1 + Math.max(leftDepth, rightDepth )
}

This can be further simplified as:

function postOrder(node){
if(!node) return 0
return 1 + Math.max(postOrder(node.left), postOrder(node.right))
}

Now let's calculate the Big O Notation:

1. Time Complexity:
2. O(N), where N is the total number of nodes in the array.
3. Space Complexity:
4. O(1), where we are not defining any space right?
5. Wrong!! why?
6. Recursion uses stack space.
7. If the tree is skewed, the recursion depth will be equal to N.
8. For a balanced binary tree, the stack depth is log⁡(N)
9. Thus the space complexity is O(H), where H is the height of the binary tree.

Potential Follow Up Question To Answer:

1. How does your solution ensure correctness for all types of binary trees (balanced, unbalanced, empty)?
2. If you were to calculate the total number of nodes in the binary tree alongside its maximum depth, how would you modify your code?
3. Could you modify your function to also return the longest path (list of nodes) in addition to the maximum depth?
4. How would you adapt your solution to work for an n-ary tree (where nodes can have more than two children)?
5. In practical applications, why might calculating the maximum depth of a tree be important?
6. How would you adapt your solution to detect if the tree contains a cycle?
7. If a tree is represented as an array (heap-style), how would you calculate the maximum depth?
8. How does your solution compare to a purely iterative BFS approach in terms of performance and simplicity?
9. Can your solution be optimized for parallel or distributed computing? If yes, how?
10. If the nodes have weights (e.g., a "cost"), how would you calculate the depth based on the maximum weight path?

Let me know what you think about the answer to these question?

#blind75 #tree #traversal