Formatted question description: https://leetcode.ca/all/1161.html

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: [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.

Note:

The number of nodes in the given tree is between 1 and 10^4.
-10^5 <= node.val <= 10^5

Companies:
Google

Related Topics:
Graph

Solution 1. Level-order Traversal

// OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        if (!root) return 0;
        int maxSum = 0, ans = 0, lv = 1;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size(), sum = 0;
            while (cnt--) {
                auto node = q.front();
                q.pop();
                sum += node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            if (sum > maxSum) {
                maxSum = sum;
                ans = lv;
            }
            ++lv;
        }
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxLevelSum(TreeNode root) {
        if (root == null)
            return 0;
        int maxSum = 0;
        int maxSumLevel = 0;
        int level = 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            level++;
            int curSum = 0;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                curSum += node.val;
                TreeNode left = node.left, right = node.right;
                if (left != null)
                    queue.offer(left);
                if (right != null)
                    queue.offer(right);
            }
            if (curSum > maxSum) {
                maxSum = curSum;
                maxSumLevel = level;
            }
        }
        return maxSumLevel;
    }
}

############

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int maxLevelSum(TreeNode root) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        int mx = Integer.MIN_VALUE;
        int i = 0;
        int ans = 0;
        while (!q.isEmpty()) {
            ++i;
            int s = 0;
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.pollFirst();
                s += node.val;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            if (mx < s) {
                mx = s;
                ans = i;
            }
        }
        return ans;
    }
}

// OJ: https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int maxLevelSum(TreeNode* root) {
        if (!root) return 0;
        int maxSum = 0, ans = 0, lv = 1;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size(), sum = 0;
            while (cnt--) {
                auto node = q.front();
                q.pop();
                sum += node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            if (sum > maxSum) {
                maxSum = sum;
                ans = lv;
            }
            ++lv;
        }
        return ans;
    }
};

# 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:
        q = deque([root])
        mx = -inf
        i = 0
        while q:
            i += 1
            s = 0
            for _ in range(len(q)):
                node = q.popleft()
                s += node.val
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            if mx < s:
                mx = s
                ans = i
        return ans

############

# 1161. Maximum Level Sum of a Binary Tree
# https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/

# 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: TreeNode) -> int:
        level = res = 1
        mmax = float('-inf')
        q = collections.deque([root])
        
        while q:
            val = 0
            n = len(q)
            
            for _ in range(n):
                node = q.popleft()
                val += node.val
                for leaf in (node.left, node.right):
                    if leaf:
                        q.append(leaf)
            
            if val > mmax:
                res = level
                mmax = val

            level += 1
        
        return res

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxLevelSum(root *TreeNode) int {
	q := []*TreeNode{root}
	mx := -0x3f3f3f3f
	i := 0
	ans := 0
	for len(q) > 0 {
		i++
		s := 0
		for n := len(q); n > 0; n-- {
			root = q[0]
			q = q[1:]
			s += root.Val
			if root.Left != nil {
				q = append(q, root.Left)
			}
			if root.Right != nil {
				q = append(q, root.Right)
			}
		}
		if mx < s {
			mx = s
			ans = i
		}
	}
	return ans
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function maxLevelSum(root: TreeNode | null): number {
    const queue = [root];
    let res = 1;
    let max = -Infinity;
    let h = 1;
    while (queue.length !== 0) {
        const n = queue.length;
        let sum = 0;
        for (let i = 0; i < n; i++) {
            const { val, left, right } = queue.shift();
            sum += val;
            left && queue.push(left);
            right && queue.push(right);
        }
        if (sum > max) {
            max = sum;
            res = h;
        }
        h++;
    }
    return res;
}

1161 - Maximum Level Sum of a Binary Tree

1161. Maximum Level Sum of a Binary Tree (Medium)

Solution 1. Level-order Traversal

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1161. Maximum Level Sum of a Binary Tree (Medium)

Solution 1. Level-order Traversal

All Problems

All Solutions