250. Count Univalue Subtrees

题目：

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.

链接： http://leetcode.com/problems/count-univalue-subtrees/

题解：

求Uni-value subtree的个数。对树来说这求count的首先思路就是递归了，不过这里要另外构造一个辅助函数来判断root为顶点的subtree是否是Uni-value subtree，假如是Uni-value的subtree，则结果可以+1，否则，我们返回递归求出的左子树的count和右节点的count。假如是Python的话可以一边计算一边返回。Java写起来可能稍微麻烦点。

Time Complexity - O(n)， Space Complexity - O(1).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countUnivalSubtrees(TreeNode root) {
        if(root == null)
            return 0;
        if(root.left == null && root.right == null)
            return 1;
        if(root.left == null) {
            if(isUniValueSubtree(root))
                return countUnivalSubtrees(root.right) + 1;
            else
                return countUnivalSubtrees(root.right);
        } else if (root.right == null) {
            if(isUniValueSubtree(root))
                return countUnivalSubtrees(root.left) + 1;
            else
                return countUnivalSubtrees(root.left);
        } else {
            if(isUniValueSubtree(root))
                return countUnivalSubtrees(root.left) + countUnivalSubtrees(root.right) + 1;
            else
                return countUnivalSubtrees(root.left) + countUnivalSubtrees(root.right);
        }
    }

    private boolean isUniValueSubtree(TreeNode root) {
        if(root == null)
            return true;
        if(root.left == null && root.right == null)
            return true;
        else if (root.left != null && root.right != null) {
            if(root.val == root.left.val && root.val == root.right.val)
                return isUniValueSubtree(root.left) && isUniValueSubtree(root.right);
            else
                return false;
        } else if (root.left == null) {
            if(root.right.val == root.val)
                return isUniValueSubtree(root.right);
            else
                return false;
        } else {
            if(root.left.val == root.val)
                return isUniValueSubtree(root.left);
            else
                return false;
        }
    }
}

Update: 优化一下

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countUnivalSubtrees(TreeNode root) {
        if(root == null)
            return 0;
        if(root.left == null && root.right == null)
            return 1;
        int res = countUnivalSubtrees(root.left) + countUnivalSubtrees(root.right) ;
        return isUniValueSubtree(root) ? res + 1 : res;
    }

    private boolean isUniValueSubtree(TreeNode root) {
        if(root == null)
            return true;
        if(root.left == null && root.right == null)
            return true;
        if(isUniValueSubtree(root.left) && isUniValueSubtree(root.right)) {
            if(root.left != null && root.right != null)
                return (root.left.val == root.right.val) && (root.left.val == root.val);
            else if(root.left != null)
                return root.left.val == root.val;
            else
                return root.right.val == root.val;
        }
        return false;
    }
}

Update: 再优化一下。注意判断两个子树是否都为Uni-value subtree的时候用了 ‘&‘，这样才能判断两个条件，否则第一个条件为false时就不判断第二个了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int count = 0;
    public int countUnivalSubtrees(TreeNode root) {
        if(root == null)
            return 0;
        isUniValueSubtree(root);
        return count;
    }

    private boolean isUniValueSubtree(TreeNode root) {
        if(root == null)
            return true;

        if(isUniValueSubtree(root.left) & isUniValueSubtree(root.right)) {
            if(root.left != null && root.left.val != root.val)
                return false;
            if(root.right != null && root.right.val != root.val)
                return false;
            count++;
            return true;
        }
        return false;
    }
}

题外话：

这道题题号是250，马克一下。

Reference:

https://leetcode.com/discuss/66805/my-java-solution-using-a-boolean-helper-function

https://leetcode.com/discuss/68057/very-easy-java-solution-post-order-recursion

https://leetcode.com/discuss/63286/7-line-accepted-code-in-java

https://leetcode.com/discuss/52210/c-one-pass-recursive-solution

https://leetcode.com/discuss/50241/recursive-java-solution-with-explanation

https://leetcode.com/discuss/50357/my-concise-java-solution

https://leetcode.com/discuss/50420/java-11-lines-added

https://leetcode.com/discuss/68057/very-easy-java-solution-post-order-recursion

https://github.com/google/google-java-format

https://leetcode.com/discuss/55213/my-ac-java-code

https://leetcode.com/discuss/53431/java-solution-with-dfs

https://leetcode.com/discuss/50452/ac-clean-java-solution