当前顶点作为拐点时，求左子树加上右子树的高度可以求出该通过该顶点的直径大小，再对该顶点和左右子节点作为拐点时直径大小进行比对，返回最大值

缺点是递归了多次

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {

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

    public int diameterOfBinaryTree(TreeNode root) {
        if(root==null)
            return 0;
        int leftHeight = dfs(root.left);
        int rightHeigh = dfs(root.right);
        return Math.max(leftHeight+rightHeigh,Math.max(diameterOfBinaryTree(root.left),diameterOfBinaryTree(root.right)));
    }
}