Given a binary tree, return the sum of values of nodes with even-valued grandparent. (A grandparent of a node is the parent of its parent, if it exists.)
If there are no nodes with an even-valued grandparent, return 0
.
Example 1:
Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5] Output: 18 Explanation: The red nodes are the nodes with even-value grandparent while the blue nodes are the even-value grandparents.
Constraints:
- The number of nodes in the tree is between
1
and10^4
. - The value of nodes is between
1
and100
.
题目大意:计算树中所有祖父结点值为偶数的结点值的和。
方法一:直接用深搜的递归做法:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 private: 14 int dfs(TreeNode *root, TreeNode *parent, TreeNode *grandParent) { 15 if (root == nullptr) return 0; 16 int sum = 0; 17 if ((grandParent != nullptr) && ((grandParent->val & 1) == 0)) 18 sum = root->val; 19 return sum + dfs(root->left, root, parent) + dfs(root->right, root, parent); 20 } 21 public: 22 int sumEvenGrandparent(TreeNode* root) { 23 if (root == nullptr) 24 return 0; 25 return dfs(root, nullptr, nullptr); 26 } 27 };