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515. Find Largest Value in Each Tree Row

Description

Given the root of a binary tree, return an array of the largest value in each row of the tree (0-indexed).

 

Example 1:

Input: root = [1,3,2,5,3,null,9]
Output: [1,3,9]

Example 2:

Input: root = [1,2,3]
Output: [1,3]

 

Constraints:

  • The number of nodes in the tree will be in the range [0, 104].
  • -231 <= Node.val <= 231 - 1

Solutions

  • /**
 * 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 List<Integer> largestValues(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int t = q.peek().val;
            for (int i = q.size(); i > 0; --i) {
                TreeNode node = q.poll();
                t = Math.max(t, node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            ans.add(t);
        }
        return ans;
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        if (!root) return {};
        queue<TreeNode*> q{ {root} };
        vector<int> ans;
        while (!q.empty()) {
            int t = q.front()->val;
            for (int i = q.size(); i; --i) {
                TreeNode* node = q.front();
                t = max(t, node->val);
                q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            ans.push_back(t);
        }
        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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
        if root is None:
            return []
        q = deque([root])
        ans = []
        while q:
            t = -inf
            for _ in range(len(q)):
                node = q.popleft()
                t = max(t, node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func largestValues(root *TreeNode) []int {
	var ans []int
	if root == nil {
		return ans
	}
	q := []*TreeNode{root}
	for len(q) > 0 {
		t := q[0].Val
		for i := len(q); i > 0; i-- {
			node := q[0]
			q = q[1:]
			t = max(t, node.Val)
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
		ans = append(ans, t)
	}
	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 largestValues(root: TreeNode | null): number[] {
    const res: number[] = [];
    const queue: TreeNode[] = [];
    if (root) {
        queue.push(root);
    }
    while (queue.length) {
        const n = queue.length;
        let max = -Infinity;
        for (let i = 0; i < n; i++) {
            const { val, left, right } = queue.shift();
            max = Math.max(max, val);
            left && queue.push(left);
            right && queue.push(right);
        }
        res.push(max);
    }
    return res;
}


  • // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
    pub fn largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = Vec::new();
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root.clone());
        }
        while !queue.is_empty() {
            let mut max = i32::MIN;
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap();
                let node = node.as_ref().unwrap().borrow();
                max = max.max(node.val);
                if node.left.is_some() {
                    queue.push_back(node.left.clone());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.clone());
                }
            }
            res.push(max);
        }
        res
    }
}

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