# 543. Diameter of Binary Tree (Easy)

https://leetcode.com/problems/diameter-of-binary-tree/

Given a binary tree, you need to compute 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.

Example:
Given a binary tree

```          1
/ \
2   3
/ \
4   5
```

Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

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

## Solutions

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int diameterOfBinaryTree(TreeNode root) {
// track keeps the depth of longest children branches for each node
Map<TreeNode, Integer> track = new HashMap<>();

recurse(root, 1, track);

int ans = 0;
for (TreeNode node : track.keySet()) {
ans = Math.max(track.get(node), ans);
}

// depth == edge, so return depth directly
return ans;
}

private int recurse(TreeNode root, int depth, Map<TreeNode, Integer> track) {
// root is null means for current path, the longest depth is depth - 1
if (root == null) {
return depth - 1;
}

// depth of left branch
int ld = recurse(root.left, depth + 1, track);

// depth of right branch
int rd = recurse(root.right, depth + 1, track);

// For each node, the diameter is the sum of left + right branches.
track.put(root, ld - depth + rd - depth);

// return the max length branch as the longest child branch of parent branch
return Math.max(ld, rd);
}
}
``````