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1315. Sum of Nodes with Even-Valued Grandparent

Description

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,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.

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

Solutions

DFS.

  • /**
 * 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);
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int res;

    int sumEvenGrandparent(TreeNode* root) {
        res = 0;
        dfs(root, root->left);
        dfs(root, root->right);
        return res;
    }

    void dfs(TreeNode* g, TreeNode* p) {
        if (!p) return;
        if (g->val % 2 == 0) {
            if (p->left) res += p->left->val;
            if (p->right) res += p->right->val;
        }
        dfs(p, p->left);
        dfs(p, p->right);
    }
};

  • # 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)
}

  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function sumEvenGrandparent(root: TreeNode | null): number {
    const dfs = (root: TreeNode | null, x: number): number => {
        if (!root) {
            return 0;
        }
        const { val, left, right } = root;
        let ans = dfs(left, val) + dfs(right, val);
        if (x % 2 === 0) {
            ans += left?.val ?? 0;
            ans += right?.val ?? 0;
        }
        return ans;
    };
    return dfs(root, 1);
}

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