1161. Maximum Level Sum of a Binary Tree

Medium

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

解題

Runtime: 143 ms, faster than 84.33%

Memory Usage: 8.7 MB, less than 55.56%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxLevelSum(root *TreeNode) int {
    m := make(map[int]int)

    var helper func(*TreeNode, int)
    helper = func(root *TreeNode, depth int) {
        if root == nil {
            return 
        }

        if _, ok := m[depth]; ok {
            m[depth] += root.Val
        } else {
            m[depth] = root.Val
        }

        helper(root.Left, depth + 1)
        helper(root.Right, depth + 1)
    }   

    helper(root, 1)

    max := 1
    for k := range m {
        if m[k] > m[max] {
            max = k
        }
    }

    return max 
}