250.Count Univalue Subtrees

Tags: Medium Tree

Links: https://leetcode-cn.com/problems/count-univalue-subtrees/

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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.

Example :

Input:  root = [5,1,5,5,5,null,5]

              5
             / \
            1   5
           / \   \
          5   5   5

Output: 4

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int countUnivalSubtrees(TreeNode* root) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return 0;
        queue<TreeNode *> q;
        q.push(root);
        int cnt = 0;

        while (!q.empty()) {
            TreeNode *tmp = q.front(); q.pop();
            //如果是叶节点，直接+1
            if (!tmp -> left && !tmp -> right) {
                ++cnt;
            }
            else if (tmp -> left && tmp -> right) {
                q.push(tmp -> left); q.push(tmp -> right);

                int target = tmp -> val;
                if (check(tmp -> left, target) && check(tmp -> right, target)) ++cnt;
            }
            else if (tmp -> left && !tmp -> right) {
                q.push(tmp -> left);
                if (check(tmp -> left, tmp -> val)) ++cnt;
            }
            else {
                q.push(tmp -> right);
                if (check(tmp -> right, tmp -> val)) ++cnt;
            }
        }

        return cnt;
    }

    bool check(TreeNode *root, int target)
    {
        queue<TreeNode *> q;
        q.push(root);
        while (!q.empty()) {
            TreeNode *tmp = q.front(); q.pop();
            if (tmp -> val != target) return false;

            if (tmp -> left) q.push(tmp -> left);
            if (tmp -> right) q.push(tmp -> right);
        }

        return true;
    }
};

思路是正确的，对每个节点去判断以此节点为根的子树的情况。注意对子树的理解，叶节点肯定可以满足，另外子树的值和根节点的值也必须相同。只不过上面的代码不够精简，所以先精简代码然后在优化。

class Solution {
public:
    int res = 0;
    int countUnivalSubtrees(TreeNode* root) {
        if (!root) return res;
        if (isUnival(root, root -> val)) ++res;
        countUnivalSubtrees(root -> left);
        countUnivalSubtrees(root -> right);
        return res;
    }
    bool isUnival(TreeNode *root, int val) {
        if (!root) return true;
        return root -> val == val && isUnival(root -> left, val) && isUnival(root -> right, val);
    }
};