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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
• C++
• Python
• Go

• /**
   * 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;
      }
  }
  
  ############
  
  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode() {}
   *     TreeNode(int val) { this.val = val; }
   *     TreeNode(int val, TreeNode left, TreeNode right) {
   *         this.val = val;
   *         this.left = left;
   *         this.right = right;
   *     }
   * }
   */
  class Solution {
      private int res;
  
      public int sumEvenGrandparent(TreeNode root) {
          res = 0;
          dfs(root, root.left);
          dfs(root, root.right);
          return res;
      }
  
      private void dfs(TreeNode g, TreeNode p) {
          if (p == null) {
              return;
          }
          if (g.val % 2 == 0) {
              if (p.left != null) {
                  res += p.left.val;
              }
              if (p.right != null) {
                  res += p.right.val;
              }
          }
          dfs(p, p.left);
          dfs(p, p.right);
      }
  }
  

• // 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;
      }
  };
  

• # Definition for a binary tree node.
  # class TreeNode:
  #     def __init__(self, val=0, left=None, right=None):
  #         self.val = val
  #         self.left = left
  #         self.right = right
  class Solution:
      def sumEvenGrandparent(self, root: TreeNode) -> int:
          self.res = 0
  
          def dfs(g, p):
              if p is None:
                  return
              if g.val % 2 == 0:
                  if p.left:
                      self.res += p.left.val
                  if p.right:
                      self.res += p.right.val
              dfs(p, p.left)
              dfs(p, p.right)
  
          dfs(root, root.left)
          dfs(root, root.right)
          return self.res
  
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  
  var res int
  
  func sumEvenGrandparent(root *TreeNode) int {
  	res = 0
  	dfs(root, root.Left)
  	dfs(root, root.Right)
  	return res
  }
  
  func dfs(g, p *TreeNode) {
  	if p == nil {
  		return
  	}
  	if g.Val%2 == 0 {
  		if p.Left != nil {
  			res += p.Left.Val
  		}
  		if p.Right != nil {
  			res += p.Right.Val
  		}
  	}
  	dfs(p, p.Left)
  	dfs(p, p.Right)
  }
  

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