1161. Maximum Level Sum of a Binary Tree

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

这道题说是给了一棵二叉树,根结点是层数1,其子结点是层数2,依次类推,让找到最小的层数,使得该层上的结点值之和最大。这里所谓的最小的层数,实际上就是说当层结点值之和相同的时候,取层数较小的层。说白了这道题的难点就是如何求每一层的结点值之和,肯定需要遍历整个二叉树,用什么样的遍历方式呢,当然是用层序最方便啦,LeetCode 中有考察层序遍历的题,Binary Tree Level Order Traversal
Binary Tree Level Order Traversal II,做过这两道的话,这道题也就没啥难度了。这里还是用 queue 来辅助遍历,先将根结点放入队列,然后开始 while 循环遍历,先将初始化为0的 level 自增1,因为根结点是层数1。因为当前 queue 里的所有结点都属于同一层,需要一起遍历完,为了防止后来新加入 queue 的结点污染,这里用个 for 循环遍历当前层的所有结点,注意 q.size() 必须放在初始化里,而不能是判断条件里,因为其会改变,累加每层所有的结点值之和到 sum,然后跟全局的 mx 对比,若 sum 大于 mx,则更新 sum 为 mx,同时更新 res 为当前层数,参见代码如下:

解法一:

class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        int res = 1, mx = root->val, level = 0;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            ++level;
            int sum = 0;
            for (int i = q.size(); i > 0; --i) {
                TreeNode* t = q.front(); q.pop();
                sum += t->val;
                if (t->left) q.push(t->left);
                if (t->right) q.push(t->right);
            }
            if (mx < sum) {
                mx = sum;
                res = level;
            }
        }
        return res;
    }
};

我们也可以强行用递归来做,由于层数可能一直会变,所以用一个数组 sums 来记录每一层的结点值之和。在递归函数中,若结点为空,直接返回。否则看若 sums 的大小于 level,说明应该增加 sums 的大小,用个 resize 来增加。然后就把当前结点值加入到 sums[level-1] 中,再分别对左右子结点调用递归函数即可。sums 数组更新好了,需要找到其中的最大值的位置,这里使用了一些 STL 内置的函数来做,用 max_element 来找最大值,用 distance 来求和第一个数字之间的距离,最后再加1即可,参见代码如下:

解法二:

class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        vector<int> sums;
        helper(root, 1, sums);
        return distance(begin(sums), max_element(begin(sums), end(sums))) + 1;
    }
    void helper(TreeNode* node, int level, vector<int>& sums) {
        if (!node) return;
        if (sums.size() < level) sums.resize(level);
        sums[level - 1] += node->val;
        helper(node->left, level + 1, sums);
        helper(node->right, level + 1, sums);
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1161

类似题目:

Binary Tree Level Order Traversal

Binary Tree Level Order Traversal II

参考资料:

https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/

https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/360968/JavaPython-3-Two-codes-language%3A-BFS-level-traversal-and-DFS-level-sum.

https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/360973/C%2B%2B-track-level-sum

LeetCode All in One 题目讲解汇总(持续更新中…)


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