[LeetCode#250] Count Univalue Subtrees

Problem:

Given a binary tree, count the number of uni-value subtrees.

A Uni-value subtree means all nodes of the subtree have the same value.

For example:
Given binary tree,

5
/ \
1   5
/ \   \
5   5   5

return

4

.

Analysis:

This problem is super simple.
But, because my poor mastery of Java language, I have committed following simple mistakes.

1. use bit wise operator as and operator.
is_unival & = (root.val == root.left.val);
Line 26: error: illegal start of expression

2. Did not consider Java would automatically optimize the running of code.
For logic operator : and ("&&")
In fact:
C = A && B
If A is false, Java would not execute B. (Thus B must be a primitive value!!! Rather than any function call).
----------------------------------------------------------------------------------------------------------
private boolean helper(TreeNode root, List<Integer> ret) {
if (root == null)
return true;
boolean is_unival = true;
if (root.left != null)
is_unival = is_unival && (root.val == root.left.val);
if (root.right != null)
is_unival = is_unival && (root.val == root.right.val);
is_unival = is_unival && helper(root.left, ret);
is_unival = is_unival && helper(root.right, ret);
if (is_unival)
ret.set(0, ret.get(0)+1);
return is_unival;
}
----------------------------------------------------------------------------------------------------------
Error output:
Input:
[5,1,5,5,5,null,5]
Output:
0
Expected:
4

If is_unival is false before reach
is_unival = is_unival && helper(root.left, ret);
is_unival = is_unival && helper(root.right, ret);

Both "helper(root.left, ret)" and "helper(root.right, ret)", would not be executed.
So as to write code as
flag = (helper(root.left, ret) && helper(root.right, ret)).
The above code would cause the problem. The way to avoid such case is to write each statement seprately and use flag_n to hold the temp result.

boolean flag1 = true, flag2 = true, flag3 = true, flag4 = true;
if (root.left != null)
flag1 = (root.val == root.left.val);
if (root.right != null)
flag2 = (root.val == root.right.val);
flag3 = helper(root.left, ret);
flag4 = helper(root.right, ret);
is_unival = flag1 && flag2 && flag3 && flag4;

Note: For conjunction equation, the initial value for each element should be false rather than true.

Solution:

public class Solution {
public int countUnivalSubtrees(TreeNode root) {
if (root == null)
return 0;
List<Integer> ret = new ArrayList<Integer> ();
ret.add(0);
helper(root, ret);
return ret.get(0);
}

private boolean helper(TreeNode root, List<Integer> ret) {
if (root == null)
return true;
boolean is_unival = true;
boolean flag1 = true, flag2 = true, flag3 = true, flag4 = true;
if (root.left != null)
flag1 = (root.val == root.left.val);
if (root.right != null)
flag2 = (root.val == root.right.val);
flag3 = helper(root.left, ret);
flag4 = helper(root.right, ret);
is_unival = flag1 && flag2 && flag3 && flag4;
if (is_unival)
ret.set(0, ret.get(0)+1);
return is_unival;
}
}