文章目录
一、题目
Given the root of a binary tree, the level of its root is 1, the level of its children is 2, and so on.
Return the smallest level x such that the sum of all the values of nodes at level x is maximal.
Example 1:
Input: root = 1,7,0,7,-8,null,null
Output: 2
Explanation:
Level 1 sum = 1.
Level 2 sum = 7 + 0 = 7.
Level 3 sum = 7 + -8 = -1.
So we return the level with the maximum sum which is level 2.
Example 2:
Input: root = 989,null,10250,98693,-89388,null,null,null,-32127
Output: 2
Constraints:
The number of nodes in the tree is in the range 1, 104.
-105 <= Node.val <= 105
二、题解
cpp
/**
* 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 maxLevelSum(TreeNode* root) {
int res = 1;
int maxSum = INT_MIN;
queue<TreeNode*> q;
q.push(root);
int level = 0;
while(!q.empty()){
int size = q.size();
int sum = 0;
level++;
while(size--){
TreeNode* t = q.front();
q.pop();
sum += t->val;
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
if(sum > maxSum){
maxSum = sum;
res = level;
}
}
return res;
}
};