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leet code

538: Count Univalue Subtrees

메모리를 많이 잡아먹고, 코드가 길었음

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
struct returnVal{
    int cnt;
    bool continued;
};

class Solution {
public:
    int countUnivalSubtrees(TreeNode* root) {
        if(!root) return 0;
        returnVal rv = solve(root);
        return rv.cnt;
    }

    returnVal solve(TreeNode* node)
    {
        returnVal rv;
        rv.cnt = 0;
        rv.continued = true;
        if(!node->left && !node->right)
        {
            rv.cnt = 1;
            return rv;
        }

        if(node->left && node->right)
        {
            returnVal rv1, rv2;
            rv1 = solve(node->left);
            rv2 = solve(node->right);
            rv.cnt += rv1.cnt + rv2.cnt;
            rv.continued = rv1.continued && rv2.continued;
            if(rv.continued)
            {
                if(node->left->val != node->val || node->right->val != node->val)
                {
                    rv.continued = false;
                }else
                {
                    rv.cnt++;
                }
            }

        }else if(!node->left && node->right)
        {
            returnVal rv1;
            rv1 = solve(node->right);
            rv.cnt += rv1.cnt;
            rv.continued = rv1.continued;
            if(rv.continued)
            {
                if(node->right->val != node->val)
                {
                    rv.continued = false;
                }else
                {
                    rv.cnt++;
                }
            }
        }else if(node->left && !node->right)
        {
            returnVal rv1;
            rv1 = solve(node->left);
            rv.cnt += rv1.cnt;
            rv.continued = rv1.continued;
            if(rv.continued)
            {
                if(node->left->val != node->val)
                {
                    rv.continued = false;
                }else
                {
                    rv.cnt++;
                }
            }
        }
        return rv;
    }
};

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class Solution {
    public:
    int countUnivalSubtrees(TreeNode* root) {
        int count = 0;
        countUnivalSubtreesRecursive(root, count);
        return count;
    }

    bool countUnivalSubtreesRecursive(TreeNode* root, int& count)
    {
        if(root == NULL) return true;

        auto isLeftUnival = countUnivalSubtreesRecursive(root->left, count);
        auto isRightUnival = countUnivalSubtreesRecursive(root->right, count);

        if(isLeftUnival && isRightUnival
           && ((root->left == NULL) || root->left->val == root->val)
           && ((root->right == NULL) || root->right->val == root->val)
          )
        {
            ++count;
            return true;
        }
        return false;
    }
};
  1. 함수의 반환형: 부모 노드에서 자식의 정보를 알고 있기 때문에, bool type으로 충분
  2. count는 stack보다 전역 변수로 두는게 좋음