解法

/**
 * 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 {
    // 经过的节点数
    int ans;
    public int diameterOfBinaryTree(TreeNode root) {
        ans = 1;
        getDepth(root);
        // 边数比节点数少1
        return ans - 1;
    }

    /**
     * 获取深度，不包括当前节点
     */
    private int getDepth(TreeNode node) {
        if (node == null) {
            return 0;
        }
        // 后序遍历获取当前节点左右子树的深度
        int left = getDepth(node.left);
        int right = getDepth(node.right);
        // 节点数等于左右子树的深度加上当前节点
        ans = Math.max(left + right + 1, ans);
        return Math.max(left, right) + 1;
    }
}