[LeetCode] 1315. Sum of Nodes with Even-Valued Grandparent
作者:互联网
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
and10^4
. - The value of nodes is between
1
and100
.
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 。来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/sum-of-nodes-with-even-valued-grandparent
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这道题看似挺复杂,其实我个人感觉考察的还是对于 DFS 和递归的理解。我这里给出一个 DFS/前序遍历 的思路。题目既然问的是如果 grandParent 节点值是偶数,才把当前节点的节点值累加到结果中,那么我们在用DFS做深度遍历的时候就一定要带上 parent 节点和 grandParent 节点的某些信息。这里只要带上了这两个节点的信息,才能在遍历到当前节点的时候决定要不要加入结果集。
时间O(n)
空间O(n)
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 int res = 0; 18 19 public int sumEvenGrandparent(TreeNode root) { 20 dfs(root, null, null); 21 return res; 22 } 23 24 private void dfs(TreeNode root, TreeNode parent, TreeNode grandParent) { 25 // base case 26 if (root == null) { 27 return; 28 } 29 if (grandParent != null && grandParent.val % 2 == 0) { 30 res += root.val; 31 } 32 dfs(root.left, root, parent); 33 dfs(root.right, root, parent); 34 } 35 }
标签:Even,1315,Valued,TreeNode,val,nodes,null,root,节点 来源: https://www.cnblogs.com/cnoodle/p/14844744.html