動機

讓每個點決定自己該做什麼

Problem

Given the root of a binary tree, return the sum of values of nodes with an even-valued grandparent. If there are no nodes with an even-valued grandparent, return 0.

A grandparent of a node is the parent of its parent if it exists.

 

Example 1:

Input: root = [6,7,8,2,7,1,3,9,Leetcode,1,4,null,null,null,5]Output: 18Explanation: The red nodes are the nodes with even-value grandparent while the blue nodes are the even-value grandparents.

Example 2:

Input: root = [1]Output: 0

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • 1 <= Node.val <= 100

Sol

讓每個點決定自己該做什麼 所以把parent的資料傳下去

class Solution {
public:
    int sumEvenGrandparent(TreeNode* root, bool p=0, bool gp=0) {
        if (!root) {
            return 0;
        } else {
            int ret = gp ? root->val : 0;
            bool even = (root->val % 2 == 0);
            int left = sumEvenGrandparent(root->left, even, p);
            int right = sumEvenGrandparent(root->right, even, p);
            return ret + left + right;
        }
    }
};