LeetCode - Easy - 144. Binary Tree Preorder Traversal-蒲公英云

LeetCode - Easy - 144. Binary Tree Preorder Traversal

Topic

Stack
Tree

Description

https://leetcode.com/problems/binary-tree-preorder-traversal/

Given the root of a binary tree, return the preorder traversal of its nodes’ values.

Example 1:

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

Example 2:

Input: root = []
Output: []

Example 3:

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

Example 4:

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

Example 5:

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

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

Follow up: Recursive solution is trivial, could you do it iteratively?

Analysis

方法一：递归法。

方法二：迭代法。

Submission

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import com.lun.util.BinaryTree.TreeNode;
public class BinaryTreePreorderTraversal { 
    public List<Integer> preorderTraversal(TreeNode root) { 
        List<Integer> result = new ArrayList<>();
        preorderTraversal(root, result);
        return result;
    }
    private void preorderTraversal(TreeNode root, List<Integer> result){ 
        if(root == null) return;
        result.add(root.val);
        preorderTraversal(root.left, result);
        preorderTraversal(root.right, result);
    }
    public List<Integer> preorderTraversal2(TreeNode root) { 
        List<Integer> result = new ArrayList<>();
        LinkedList<TreeNode> stack = new LinkedList<>();
        if(root != null) stack.push(root);
        while(!stack.isEmpty()) { 
            TreeNode node = stack.pop();
            result.add(node.val);
            if(node.right != null)
                stack.push(node.right);
            if(node.left != null)
                stack.push(node.left);
        }
        return result;
    }
}

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 BinaryTreePreorderTraversalTest { 
    @Test
    public void test() { 
        BinaryTreePreorderTraversal obj = new BinaryTreePreorderTraversal();
        assertThat(obj.preorderTraversal(BinaryTree.integers2BinaryTree(1, null, 2, 3)), //
                IsIterableContainingInOrder.contains(1, 2, 3));
        assertThat(obj.preorderTraversal(null), IsEmptyCollection.empty());
        assertThat(obj.preorderTraversal(BinaryTree.integers2BinaryTree(1)), //
                IsIterableContainingInOrder.contains(1));
        assertThat(obj.preorderTraversal(BinaryTree.integers2BinaryTree(1,2)), //
                IsIterableContainingInOrder.contains(1,2));
        assertThat(obj.preorderTraversal(BinaryTree.integers2BinaryTree(1,null,2)), //
                IsIterableContainingInOrder.contains(1,2));
    }
    @Test
    public void test2() { 
        BinaryTreePreorderTraversal obj = new BinaryTreePreorderTraversal();
        assertThat(obj.preorderTraversal2(BinaryTree.integers2BinaryTree(1, null, 2, 3)), //
                IsIterableContainingInOrder.contains(1, 2, 3));
        assertThat(obj.preorderTraversal2(null), IsEmptyCollection.empty());
        assertThat(obj.preorderTraversal2(BinaryTree.integers2BinaryTree(1)), //
                IsIterableContainingInOrder.contains(1));
        assertThat(obj.preorderTraversal2(BinaryTree.integers2BinaryTree(1,2)), //
                IsIterableContainingInOrder.contains(1,2));
        assertThat(obj.preorderTraversal2(BinaryTree.integers2BinaryTree(1,null,2)), //
                IsIterableContainingInOrder.contains(1,2));
    }
}