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
Solution 1: DFS
We design a function dfs(root, x), which represents the sum of the values of the nodes that meet the conditions in the subtree with root as the root node and x as the value of the parent node of root. The answer is dfs(root, 1).
The execution process of the function dfs(root, x) is as follows:
- If root is null, return 0.
- Otherwise, we recursively calculate the answers of the left and right subtrees of root, that is, dfs(root.left, root.val) and dfs(root.right, root.val), and add them to the answer. If x is even, we check whether the left and right children of root exist. If they exist, we add their values to the answer.
- Finally, return the answer.
The time complexity is O(n), and the space complexity is O(n). Where n is the number of nodes.
PythonJavaC++GoTypeScript
# 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:
def dfs(root: TreeNode, x: int) -> int:
if root is None:
return 0
ans = dfs(root.left, root.val) + dfs(root.right, root.val)
if x % 2 == 0:
if root.left:
ans += root.left.val
if root.right:
ans += root.right.val
return ans
return dfs(root, 1)(code-box)
/**
* 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 {
public int sumEvenGrandparent(TreeNode root) {
return dfs(root, 1);
}
private int dfs(TreeNode root, int x) {
if (root == null) {
return 0;
}
int ans = dfs(root.left, root.val) + dfs(root.right, root.val);
if (x % 2 == 0) {
if (root.left != null) {
ans += root.left.val;
}
if (root.right != null) {
ans += root.right.val;
}
}
return ans;
}
}(code-box)
/**
* 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 sumEvenGrandparent(TreeNode* root) {
function<int(TreeNode*, int)> dfs = [&](TreeNode* root, int x) {
if (!root) {
return 0;
}
int ans = dfs(root->left, root->val) + dfs(root->right, root->val);
if (x % 2 == 0) {
if (root->left) {
ans += root->left->val;
}
if (root->right) {
ans += root->right->val;
}
}
return ans;
};
return dfs(root, 1);
}
};(code-box)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func sumEvenGrandparent(root *TreeNode) int {
var dfs func(*TreeNode, int) int
dfs = func(root *TreeNode, x int) int {
if root == nil {
return 0
}
ans := dfs(root.Left, root.Val) + dfs(root.Right, root.Val)
if x%2 == 0 {
if root.Left != nil {
ans += root.Left.Val
}
if root.Right != nil {
ans += root.Right.Val
}
}
return ans
}
return dfs(root, 1)
}(code-box)
/**
* 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);
}(code-box)
