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# Diameter of Binary Tree LeetCode Solution

## Problem – Diameter of Binary Tree

Given the `root` of a binary tree, return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.

The length of a path between two nodes is represented by the number of edges between them.

Example 1:

``````Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].``````

Example 2:

``````Input: root = [1,2]
Output: 1``````

Constraints:

• The number of nodes in the tree is in the range `[1, 104]`.
• `-100 <= Node.val <= 100`

### Diameter of Binary Tree LeetCode Solution in Java

``````public class Solution {
int max = 0;

public int diameterOfBinaryTree(TreeNode root) {
maxDepth(root);
return max;
}

private int maxDepth(TreeNode root) {
if (root == null) return 0;

int left = maxDepth(root.left);
int right = maxDepth(root.right);

max = Math.max(max, left + right);

return Math.max(left, right) + 1;
}
}``````

### Diameter of Binary Tree LeetCode Solution in C++

``````int diameterOfBinaryTree(TreeNode* root) {
int d=0;
rec(root, d);
return d;
}

int rec(TreeNode* root, int &d) {
if(root == NULL) return 0;
int ld = rec(root->left, d);
int rd = rec(root->right, d);
d=max(d,ld+rd);
return max(ld,rd)+1;
}

``````

### Diameter of Binary Tree LeetCode Solution in Python

``````class Solution(object):
def diameterOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.ans = 0

def depth(p):
if not p: return 0
left, right = depth(p.left), depth(p.right)
self.ans = max(self.ans, left+right)
return 1 + max(left, right)

depth(root)
return self.ans``````
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