250. Count Univalue Subtrees

三刷

100.00 %
class Solution {
    public int countUnivalSubtrees(TreeNode root) {
        int[] count = new int[1];
        helper(root, count);
        return count[0];
    }
    // Add current subtree to result
    // Return if current subtree is a univalsubtree
    private boolean helper(TreeNode node, int[] count) {
        if (node == null) {
            return true;
        }
        if (node.left == null && node.right == null) {
            count[0]++;
            return true;
        }
        boolean left = helper(node.left, count);
        boolean right = helper(node.right, count);
        if ((node.left == null || node.left.val == node.val && left)
           && (node.right == null || node.right.val == node.val && right)) {
            count[0]++;
            return true;
        }
        return false;
    }
}


Version #2(better)
12.05 % 

public class Solution {
    private int count;
    public int countUnivalSubtrees(TreeNode root) {
        //current node is the root of a UnivalSubtree:
        //left &  right children are (null || UnivalSubtrees) & value equals to current node
        //return boolean -> is not UnivalSubtree
        if (root == null) return 0;
        count = 0;
        countUnivalSubtreesHelper(root);
        return count;
    }
    private boolean countUnivalSubtreesHelper(TreeNode root) {
        if (root == null) return true;
        boolean left = countUnivalSubtreesHelper(root.left);
        boolean right = countUnivalSubtreesHelper(root.right);
        if (left && right) {
            if ((root.left == null || root.left.val == root.val) && (root.right == null || root.right.val == root.val)) {
                count += 1;
                return true;
            }
        }
        return false;
    }
}

Version #1
对leaf的判断写的不够简洁，其实这道题直接把base case交给root == null就可以了
更新版看Version #2
12.05 % 
public class Solution {
    private int count;
    public int countUnivalSubtrees(TreeNode root) {
        //current node is the root of a UnivalSubtree:
        //left &  right children are (null || UnivalSubtrees) & value equals to current node
        //return boolean -> is not UnivalSubtree
        if (root == null) return 0;
        count = 0;
        countUnivalSubtreesHelper(root);
        return count;
    }
    private boolean countUnivalSubtreesHelper(TreeNode root) {
        if (root.left == null && root.right == null) {
            count += 1;
            return true;
        }
        boolean left = root.left == null ? true : countUnivalSubtreesHelper(root.left);
        boolean right = root.right == null ? true : countUnivalSubtreesHelper(root.right);
        if (left && right) {
            if ((root.left == null || root.left.val == root.val) && (root.right == null || root.right.val == root.val)) {
                count += 1;
                return true;
            }
        }
        return false;
    }
}