Formatted question description: https://leetcode.ca/all/1315.html

1315. Sum of Nodes with Even-Valued Grandparent (Medium)

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 and 10^4.
  • The value of nodes is between 1 and 100.

Related Topics:
Tree, Depth-first Search

Solution 1. Pre-order Traversal

// OJ: https://leetcode.com/problems/sum-of-nodes-with-even-valued-grandparent/
// Time: O(N)
// Space: O(N)
class Solution {
    int ans = 0;
    vector<TreeNode*> v;
    void preorder(TreeNode *root, int level) {
        if (!root) return;
        if (v.size() <= level) v.push_back(root);
        else v[level] = root;
        if (level - 2 >= 0 && v[level - 2]->val % 2 == 0) ans += root->val;
        preorder(root->left, level + 1);
        preorder(root->right, level + 1);
    }
public:
    int sumEvenGrandparent(TreeNode* root) {
        preorder(root, 0);
        return ans;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public int sumEvenGrandparent(TreeNode root) {
            int sum = 0;
            Queue<TreeNode> nodeQueue = new LinkedList<TreeNode>();
            Queue<Boolean> booleanQueue = new LinkedList<Boolean>();
            nodeQueue.offer(root);
            booleanQueue.offer(false);
            while (!nodeQueue.isEmpty()) {
                TreeNode node = nodeQueue.poll();
                boolean isEven = booleanQueue.poll();
                boolean curEven = node.val % 2 == 0;
                TreeNode left = node.left, right = node.right;
                if (left != null) {
                    if (isEven)
                        sum += left.val;
                    nodeQueue.offer(left);
                    booleanQueue.offer(curEven);
                }
                if (right != null) {
                    if (isEven)
                        sum += right.val;
                    nodeQueue.offer(right);
                    booleanQueue.offer(curEven);
                }
            }
            return sum;
        }
    }
    
  • // OJ: https://leetcode.com/problems/sum-of-nodes-with-even-valued-grandparent/
    // Time: O(N)
    // Space: O(N)
    class Solution {
        int ans = 0;
        vector<TreeNode*> v;
        void preorder(TreeNode *root, int level) {
            if (!root) return;
            if (v.size() <= level) v.push_back(root);
            else v[level] = root;
            if (level - 2 >= 0 && v[level - 2]->val % 2 == 0) ans += root->val;
            preorder(root->left, level + 1);
            preorder(root->right, level + 1);
        }
    public:
        int sumEvenGrandparent(TreeNode* root) {
            preorder(root, 0);
            return ans;
        }
    };
    
  • print("Todo!")
    

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