1302. Deepest Leaves Sum
Medium
Given the root of a binary tree, return the sum of values of its deepest leaves.
Example 1:
Input: root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
Output: 15
Example 2:
Input: root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
Output: 19
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. 1 <= Node.val <= 100
思路
（略）
/**
* 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 deepestLeavesSum(TreeNode* root) {
vector<TreeNode*> current_layer;
vector<TreeNode*> next_layer;
current_layer.push_back(root);
int sum = 0;
while(!current_layer.empty()) {
for (auto node: current_layer) {
if (node->left != nullptr) {
next_layer.push_back(node->left);
}
if (node->right != nullptr) {
next_layer.push_back(node->right);
}
}
if (next_layer.empty()) {
for (auto node: current_layer) {
sum += node->val;
}
break;
}
current_layer = next_layer;
next_layer.clear();
}
return sum;
}
};
Runtime: 40 ms, faster than 57.22% of C++ online submissions for Deepest Leaves Sum.
Memory Usage: 38.7 MB, less than 38.77% of C++ online submissions for Deepest Leaves Sum.