Maximum Depth of Binary Tree (LeetCode 104) | Full Solution with animations | Study Algorithms

[@Nishi-Tiwari]

You are great man! I am amazed the way you simply the problems.

[@hinocenciopaulo]

Finally, I understand this implementation. Thank you so much 🙏!!

[@pradeeps9633]

Bro i have exam for amadeous next month plz help me in preparation in coding. you're the only one who taught me algorithms clearly.plzz bro😢😢

[@ArjunKumar-xh2xo]

Thank you !!! :)

[@salmamohammed1200]

Height of Binary tree is the number of edges along the longest path from root to leaf
Max Depth of Binary tree is the number of nodes along the longest path from root to leaf
Max Depth = Height + 1
So the answer should be 3 right for the example you have given

[@nikoo28]

These terms will be used interchangeably…but you get the correct idea 👍

[@chiruchiruchiranjeevi3237]

@@nikoo28 bor even my test cases are not passing its showing wrong, please do help me in this once..

[@brinderdhaliwal3570]

@@chiruchiruchiranjeevi3237 You probably have to check for an empty tree (root == null)

[@honey-xr5kp]

all you gotta do is change the numberOfLevels to be initialized as 0 instead of -1 and it should all work. 0 for depth, -1 for height.@@chiruchiruchiranjeevi3237

[@mehnazahmad7929]

Loved your all the videos. But this code has one flaw. If you see the queue size will never be zero until it reached leaf node as we are enqueuing it simultaneously will children nodes, as a result height will never increment to its correct value.

[@nikoo28]

It is always the correct value. The code passes all cases on LeetCode. Can you elaborate more?

[@RN-jo8zt]

Height of Binary tree is the number of edges along the longest path from root to leaf
but why we adding 1 also leaf node.which is wrong correct?
def height(root):
# Base case: Empty tree has a height equals 0
if root is None:
return 0

# Calling Left and Right node recursively
return 1 + max(height(root.left), height(root.right))

[@nikoo28]

i did not understand your question. The 1 actually accounts for your count. If you do not have that..the result will always be 0

[@alekseiwe]

Thank man!

[@subee128]

Thanks

[@gauravmishra8782]

nice video nikhil sir!!!

[@startercoder]

Helpful man!

[@nikoo28]

Glad to hear it!

[@gokulnaathb2627]

how to calculate the distance between the root and each of the leaf nodes in the brute-force approach?

[@nikoo28]

you keep on traversing each direction, until you find the desired leaf node

[@niko9338]

any resource on building this basic type of binary tree? i find it harder to build it than binary search trees that split the values based on their size to left and right subtrees

[@ivanchl]

How does an TreeNode "element" represent an "int" if int value is stored in the element.val (int val) inside a TreeNode object? Are we not supposed to reference the int value with element.val?

[@nikoo28]

What do you mean? If the element.val is int, it will be of the int type.
Can you clarify your question please

[@ivanchl]

@@nikoo28 Thank you so much for replying! I will give an example from your code: if you add to queue elementQueue.add(root), isn't it supposed to be elementQueue.add(root.val)? How does it retrieve the integer from just a "root" TreeNode object if we do not reference exactly to the value(root.val)?
public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
....... etc.

[@nikoo28]

that is because if you retrieve the root, you will get a TreeNode object. you can then fetch the value using ".val"

[@chiruchiruchiranjeevi3237]

plz don't mind, this 14th and 11th from this playlist are the same right?

[@ashok2089]

Yes. His intention is to convey that the same problem can be asked in two ways: Maximum Depth of Binary Tree, Height of binary tree.

[@hydrocy.9165]

int maxDepth(TreeNode* root) {
int maxDepth = 0; // Initialize the maximum depth
int count = 0; // Initialize the current depth counter
dfs(root, count, maxDepth);
return maxDepth;
}
private:
void dfs(TreeNode* node, int count, int &maxDepth) {
if (node == NULL) return;
count++; // Increment counter to reflect current depth
if (count > maxDepth) {
maxDepth = count; // Update maximum depth
}
dfs(node->left, count, maxDepth);
dfs(node->right, count, maxDepth);
}
}; bhaiya this code works but i cant understand how every recursion call maintain its own count variable