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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