11-line C++ solution

  • 0

    isValid: decide whether a tree is a Univalue Subtree
    valid Univalue Subtree:

    1. single node
    2. two child is either empty or is also a valid Univalue Subtree AND equal to root value
    class Solution {
        int countUnivalSubtrees(TreeNode* root) {
            if(!root) return 0;
            return countUnivalSubtrees(root->left) + countUnivalSubtrees(root->right) + isValid(root);
        bool isValid(TreeNode* root)
            if(!root || (root && !root->left && !root->right)) return true;
            if((!root->left || (isValid(root->left) && root->val==root->left->val)) &&
                (!root->right || (isValid(root->right) && root->val==root->right->val))
            )return true;
            return false;

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