LeetCode - Medium - 515. Find Largest Value in Each Tree Row-蒲公英云

LeetCode - Medium - 515. Find Largest Value in Each Tree Row

Topic

Tree
Depth-first Search
Breadth-first Search

Description

https://leetcode.com/problems/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: []

Constraints:

The number of nodes in the tree will be in the range [ 0 , 1 0 4 ] [0, 10^4] [0,104].
− 2 31 < = N o d e . v a l < = 2 31 − 1 -2^{31} <= Node.val <= 2^{31} - 1 −231<=Node.val<=231−1

Analysis

方法一：BFS

方法二：DFS

Submission

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import com.lun.util.BinaryTree.TreeNode;
public class FindLargestValueInEachTreeRow { 
    //方法一：BFS
    public List<Integer> largestValues(TreeNode root) { 
        List<Integer> result = new ArrayList<>();
        if(root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) { 
            int max = Integer.MIN_VALUE;     
            for(int size = queue.size(); size > 0; --size ) { 
                TreeNode node = queue.poll();
                if(node.val > max)
                    max = node.val;
                if(node.left != null)
                    queue.offer(node.left);
                if(node.right != null)
                    queue.offer(node.right);
            }
            result.add(max);
        }
        return result;
    }
    //方法二：DFS
    public List<Integer> largestValues2(TreeNode root) { 
        List<Integer> result = new ArrayList<>();
        largestValues2(root, result, 0);
        return result;
    }
    private void largestValues2(TreeNode node, List<Integer> result, int level) { 
        if(node == null) return;
        if(level >= result.size()) { 
            result.add(node.val);
        }else { 
            if(result.get(level) < node.val)
                result.set(level, node.val);
        }
        largestValues2(node.left, result, level + 1);
        largestValues2(node.right, result, level + 1);
    }
}

Test

import static org.junit.Assert.*;
import org.hamcrest.collection.IsEmptyCollection;
import org.hamcrest.collection.IsIterableContainingInOrder;
import org.junit.Test;
import com.lun.util.BinaryTree;
public class FindLargestValueInEachTreeRowTest { 
    @Test
    public void test() { 
        FindLargestValueInEachTreeRow obj = new FindLargestValueInEachTreeRow();
        assertThat(obj.largestValues(BinaryTree.integers2BinaryTree(1, 3, 2, 5, 3, null, 9)),//
                IsIterableContainingInOrder.contains(1, 3, 9));
        assertThat(obj.largestValues(BinaryTree.integers2BinaryTree(1, 2, 3)),//
                IsIterableContainingInOrder.contains(1, 3));
        assertThat(obj.largestValues(BinaryTree.integers2BinaryTree(1)),//
                IsIterableContainingInOrder.contains(1));
        assertThat(obj.largestValues(BinaryTree.integers2BinaryTree(1, null, 2)),//
                IsIterableContainingInOrder.contains(1, 2));
        assertThat(obj.largestValues(null), IsEmptyCollection.empty());
    }
    @Test
    public void test2() { 
        FindLargestValueInEachTreeRow obj = new FindLargestValueInEachTreeRow();
        assertThat(obj.largestValues2(BinaryTree.integers2BinaryTree(1, 3, 2, 5, 3, null, 9)),//
                IsIterableContainingInOrder.contains(1, 3, 9));
        assertThat(obj.largestValues2(BinaryTree.integers2BinaryTree(1, 2, 3)),//
                IsIterableContainingInOrder.contains(1, 3));
        assertThat(obj.largestValues2(BinaryTree.integers2BinaryTree(1)),//
                IsIterableContainingInOrder.contains(1));
        assertThat(obj.largestValues2(BinaryTree.integers2BinaryTree(1, null, 2)),//
                IsIterableContainingInOrder.contains(1, 2));
        assertThat(obj.largestValues2(null), IsEmptyCollection.empty());
    }
}