My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int counter = 0;
    public int countUnivalSubtrees(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        helper(root);
        return counter;
    }
    
    private int helper(TreeNode root) {
        if (root.left == null && root.right == null) {
            counter++;
            return root.val;
        }
        else if (root.left == null) {
            int ret = helper(root.right);
            if (ret == -1) {
                return -1;
            }
            else if (ret != root.val) {
                return -1;
            }
            else {
                counter++;
                return ret;
            }
        }
        else if (root.right == null) {
            int ret = helper(root.left);
            if (ret == -1) {
                return -1;
            }
            else if (ret != root.val) {
                return -1;
            }
            else {
                counter++;
                return ret;
            }
        }
        else {
            int ret1 = helper(root.left);
            int ret2 = helper(root.right);
            if (ret1 == -1 || ret2 == -1) {
                return -1;
            }
            else {
                if (ret1 == root.val && ret2 == root.val) {
                    counter++;
                    return ret1;
                }
                else {
                    return -1;
                }
            }
        }
    }
}

post-order recursion

看了下答案，和我差不多的思路，但写的更加简洁，可能有些情况不用再细分了。

Anyway, Good luck, Richardo! — 09/06/2016