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# 543 Diameter of Binary Tree

## 543. [Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/description/)

## 1. Question

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.

## 2. Implementation

**(1) DFS Recursion**

思路: 这题其实和124找max path sum一样，基本的步骤都是先用postorder traversal 找出当前node的left path的信息，和right path的信息，然后用一个变量找出当前的信息加上左右两边信息的总和 与 当前已知的最大信息之间的较大者。返回的值则是当前的信息(节点的个数或者节点的值) 加上左右两边信息较大的一方

```java
/**
 * 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) {
        int[] diameter = new int[1];

        getDiameterOfTree(root, diameter);
        return diameter[0];
    }

    public int getDiameterOfTree(TreeNode node, int[] diameter) {
        if (node == null) {
            return 0;
        }

        int leftDiameter = getDiameterOfTree(node.left, diameter);
        int rightDiameter = getDiameterOfTree(node.right, diameter);
        diameter[0] = Math.max(diameter[0], leftDiameter + rightDiameter);
        return 1 + Math.max(leftDiameter, rightDiameter);
    }
}
```

## 3. Time & Space Complexity

**DFS Recursion:** 时间复杂度O(n), n为树node的个数, 空间复杂度O(n)