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1315. 祖父节点值为偶数的节点和:
给你一棵二叉树,请你返回满足以下条件的所有节点的值之和:
- 该节点的祖父节点的值为偶数。(一个节点的祖父节点是指该节点的父节点的父节点。)
如果不存在祖父节点值为偶数的节点,那么返回 0
。
样例 1:
输入:
root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
输出:
18
解释:
图中红色节点的祖父节点的值为偶数,蓝色节点为这些红色节点的祖父节点。
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提示:
- 树中节点的数目在 1 到 之间。
- 每个节点的值在 1 到 100 之间。
分析
- 面对这道算法题目,二当家的陷入了沉思。
- 遍历结构,一般就是使用循环或是递归,这里显然递归更加直观易懂,所以选择使用套娃递归大法。
- 二叉树的关联是单向的,无法从一个节点关联到它的祖父节点,所以我们需要把祖父节点或者说祖父节点的信息传递到递归中。
- 父亲节点将会是当前节点子节点的祖父节点,而且也无法从一个节点关联到它的父亲节点,所以父亲节点的信息也要一并传递到递归中。
题解
rust
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn sum_even_grandparent(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
fn dfs(mut node: Option<Rc<RefCell<TreeNode>>>, parentIsEven: bool, grandparentIsEven: bool) -> i32 {
match node {
Some(node) => {
let mut ans = 0;
let mut node = node.borrow_mut();
if grandparentIsEven {
ans += node.val;
}
ans += dfs(node.left.take(), node.val & 1 == 0, parentIsEven);
ans += dfs(node.right.take(), node.val & 1 == 0, parentIsEven);
ans
}
_ => { 0 }
}
}
dfs(root, false, false)
}
}
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go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func sumEvenGrandparent(root *TreeNode) int {
var dfs func(*TreeNode, bool, bool) int
dfs = func(node *TreeNode, parentIsEven bool, grandparentIsEven bool) int {
if node == nil {
return 0
}
ans := 0
if grandparentIsEven {
ans += node.Val
}
ans += dfs(node.Left, (node.Val & 1) == 0, parentIsEven)
ans += dfs(node.Right, (node.Val & 1) == 0, parentIsEven)
return ans
}
return dfs(root, false, false)
}
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typescript
/**
* 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 = (node: TreeNode, parentIsEven: Boolean, grandparentIsEven: Boolean): number => {
if (node == null) {
return 0;
}
let ans = 0;
if (grandparentIsEven) {
ans += node.val;
}
ans += dfs(node.left, (node.val & 1) == 0, parentIsEven);
ans += dfs(node.right, (node.val & 1) == 0, parentIsEven);
return ans;
};
return dfs(root, false, false);
};
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c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int dfs(struct TreeNode* node, bool parentIsEven, bool grandparentIsEven) {
if (node == NULL) {
return 0;
}
int ans = 0;
if (grandparentIsEven) {
ans += node->val;
}
ans += dfs(node->left, (node->val & 1) == 0, parentIsEven);
ans += dfs(node->right, (node->val & 1) == 0, parentIsEven);
return ans;
}
int sumEvenGrandparent(struct TreeNode* root){
return dfs(root, false, false);
}
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c++
/**
* 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 {
private:
int dfs(TreeNode* node, bool parentIsEven, bool grandparentIsEven) {
if (node == nullptr) {
return 0;
}
int ans = 0;
if (grandparentIsEven) {
ans += node->val;
}
ans += dfs(node->left, (node->val & 1) == 0, parentIsEven);
ans += dfs(node->right, (node->val & 1) == 0, parentIsEven);
return ans;
}
public:
int sumEvenGrandparent(TreeNode* root) {
return dfs(root, false, false);
}
};
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java
/**
* 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, false, false);
}
private int dfs(TreeNode node, boolean parentIsEven, boolean grandparentIsEvent) {
if (node == null) {
return 0;
}
int ans = 0;
if (grandparentIsEvent) {
ans += node.val;
}
ans += dfs(node.left, (node.val & 1) == 0, parentIsEven);
ans += dfs(node.right, (node.val & 1) == 0, parentIsEven);
return ans;
}
}
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python
# 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(node: TreeNode, parent_is_even: bool, grandparent_is_even: bool) -> int:
if node is None:
return 0
ans = 0
if grandparent_is_even:
ans += node.val
ans += dfs(node.left, (node.val & 1) == 0, parent_is_even)
ans += dfs(node.right, (node.val & 1) == 0, parent_is_even)
return ans
return dfs(root, False, False)
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原题传送门:https://leetcode.cn/problems/sum-of-nodes-with-even-valued-grandparent/
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