python3 解法, 执行用时: 264 ms, 内存消耗: 19.4 MB, 提交时间: 2022-11-29 12:11:31

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxLevelSum(self, root: Optional[TreeNode]) -> int:
        ans, maxSum = 1, root.val
        level, q = 1, [root]
        while q:
            sum, nq = 0, []
            for node in q:
                sum += node.val
                if node.left:
                    nq.append(node.left)
                if node.right:
                    nq.append(node.right)
            if sum > maxSum:
                ans, maxSum = level, sum
            q = nq
            level += 1
        return ans

golang 解法, 执行用时: 88 ms, 内存消耗: 7.9 MB, 提交时间: 2022-11-29 12:11:16

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxLevelSum(root *TreeNode) int {
    ans, maxSum := 1, root.Val
    q := []*TreeNode{root}
    for level := 1; len(q) > 0; level++ {
        tmp := q
        q = nil
        sum := 0
        for _, node := range tmp {
            sum += node.Val
            if node.Left != nil {
                q = append(q, node.Left)
            }
            if node.Right != nil {
                q = append(q, node.Right)
            }
        }
        if sum > maxSum {
            ans, maxSum = level, sum
        }
    }
    return ans
}

javascript 解法, 执行用时: 156 ms, 内存消耗: 60.6 MB, 提交时间: 2022-11-29 12:11:00

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxLevelSum = function(root) {
    let ans = 1, maxSum = root.val;
    let q = [];
    q.push(root);
    for (let level = 1; q.length > 0; ++level) {
        const nq = [];
        let sum = 0;
        for (const node of q) {
            sum += node.val;
            if (node.left) {
                nq.push(node.left);
            }
            if (node.right) {
                nq.push(node.right);
            }
        }
        if (sum > maxSum) {
            maxSum = sum;
            ans = level;
        }
        q = nq;
    }
    return ans;
};

javascript 解法, 执行用时: 152 ms, 内存消耗: 60.6 MB, 提交时间: 2022-11-29 12:10:43

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxLevelSum = function(root) {
    const sum = [];
    const dfs = (node, level) => {
        if (level === sum.length) {
            sum.push(node.val);
        } else {
            sum.splice(level, 1, sum[level] + node.val);
        }
        if (node.left) {
            dfs(node.left, level + 1);
        }
        if (node.right) {
            dfs(node.right, level + 1);
        }
    }
    dfs(root, 0);
    let ans = 0;
    for (let i = 0; i < sum.length; ++i) {
        if (sum[i] > sum[ans]) {
            ans = i;
        }
    }
    return ans + 1; // 层号从 1 开始
};

golang 解法, 执行用时: 96 ms, 内存消耗: 8.2 MB, 提交时间: 2022-11-29 12:10:26

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxLevelSum(root *TreeNode) (ans int) {
    sum := []int{}
    var dfs func(*TreeNode, int)
    dfs = func(node *TreeNode, level int) {
        if level == len(sum) {
            sum = append(sum, node.Val)
        } else {
            sum[level] += node.Val
        }
        if node.Left != nil {
            dfs(node.Left, level+1)
        }
        if node.Right != nil {
            dfs(node.Right, level+1)
        }
    }
    dfs(root, 0)
    for i, s := range sum {
        if s > sum[ans] {
            ans = i
        }
    }
    return ans + 1 // 层号从 1 开始
}

python3 解法, 执行用时: 272 ms, 内存消耗: 19.5 MB, 提交时间: 2022-11-29 12:10:03

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxLevelSum(self, root: Optional[TreeNode]) -> int:
        sum = []
        def dfs(node: TreeNode, level: int) -> None:
            if level == len(sum):
                sum.append(node.val)
            else:
                sum[level] += node.val
            if node.left:
                dfs(node.left, level + 1)
            if node.right:
                dfs(node.right, level + 1)
        dfs(root, 0)
        return sum.index(max(sum)) + 1  # 层号从 1 开始