515. Find Largest Value in Each Tree Row

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]

Example 3:

Input: root = [1]
Output: [1]

Example 4:

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

Example 5:

Input: root = []
Output: []

ex1

Constraints:

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

Solution

/**
 * 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) {
        
        Queue<TreeNode> q = new LinkedList<>();
        
        if (root != null) {
            q.add(root);
        }
        List<Integer> res = new ArrayList<>();
        while (q.size() > 0) {
            
            int size = q.size();
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                max = Math.max(node.val, max);
                
                if (node.left != null) q.add(node.left);
                if (node.right != null) q.add(node.right);
            }
            
            res.add(max);
        }
        
        return res;
    }
    
    public List<Integer> largestValues2(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        
        dfs(root, res, 0);
        
        return res;
    }
    
    
    void dfs(TreeNode node, List<Integer> res, int level) {
        if (node == null) return;
        
        if (res.size() == level) {
            res.add(node.val);
        }
        
        Integer max = Math.max(res.get(level), node.val);
        res.set(level, max);
        
        dfs(node.left, res, level + 1);
        dfs(node.right, res, level + 1);
    }
}

Solution 2021-10-20

class Solution {
    public List<Integer> largestValues(TreeNode root) {
        
        if (root == null) return Collections.emptyList();
        Queue<TreeNode> q = new ArrayDeque<>();
        List<Integer> res = new ArrayList<>();
        q.add(root);
        while (q.size() > 0) {
            int size = q.size();
            int max = Integer.MIN_VALUE;
            for (int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                max = Math.max(node.val, max);
                if (node.left != null) {
                    q.add(node.left);
                }
                if (node.right != null) {
                    q.add(node.right);
                }
            }
            res.add(max);
        }
        return res;
    }
}