思路:先遍历二叉树,为偶数的则向下找两个层加起来
class Solution { public: int an = 0; int sumEvenGrandparent(TreeNode* root) { bianli(root); return an; } void findSon(int level, TreeNode *node){ if(level == 2){ an += node->val; // cout << an << endl; return; }else{ if(node->left != nullptr){ findSon(level+1, node->left); } if(node->right != nullptr){ findSon(level+1, node->right); } } } void bianli(TreeNode *node){ if(node->val % 2 == 0){ findSon(0, node); } if(node->left != nullptr){ bianli(node->left); } if(node->right != nullptr){ bianli(node->right); } } };
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