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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:

  1. The number of nodes in the given tree is between 1 and 10^4.
  2. -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;
    }
    
    

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