[] C010_ binary tree diameter (recursive)

One, Title Description

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.

Second, the problem solution

Method One: Reflections

Precautions: The maximum diameter possible without root, so you can only use variable max to record the maximum diameter of each node as the root.

• Recurrence equation: maximum height and the maximum diameter = root node to each of the left and right subtrees.
  • $diameter = max(left + right，diameter )$

int max;
public int diameterOfBinaryTree(TreeNode root) {
  dfs(root);
  return max;
}
private int dfs(TreeNode root) {
  if (root == null)
    return 0;
  int l = dfs(root.left);
  int r = dfs(root.right);
  max = Math.max(l+r, max);
  return Math.max(l, r) + 1;
}

Complexity Analysis

• time complexity: $O (n)$ ,
• Space complexity: $O (n)$ ,

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