• leetcode 1315. Sum of Nodes with Even-Valued Grandparent


    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.

    题目大意:计算树中所有祖父结点值为偶数的结点值的和。

    方法一:直接用深搜的递归做法:

     1 /**
     2  * Definition for a binary tree node.
     3  * struct TreeNode {
     4  *     int val;
     5  *     TreeNode *left;
     6  *     TreeNode *right;
     7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
    10  * };
    11  */
    12 class Solution {
    13 private:
    14     int dfs(TreeNode *root, TreeNode *parent, TreeNode *grandParent) {
    15         if (root == nullptr) return 0;
    16         int sum = 0;
    17         if ((grandParent != nullptr) && ((grandParent->val & 1) == 0))
    18             sum = root->val;
    19         return sum + dfs(root->left, root, parent) + dfs(root->right, root, parent);
    20     }
    21 public:
    22     int sumEvenGrandparent(TreeNode* root) {
    23         if (root == nullptr)
    24             return 0;
    25         return dfs(root, nullptr, nullptr);
    26     }
    27 };
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  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/13063044.html
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