Leetcode – Binary Tree Preorder Traversal (Java)

Analysis

Preorder binary tree traversal is a classic interview problem about trees. The key to solve this problem is to understand the following:

What is preorder? (parent node is processed before its children)
Use Stack from Java Core library

The key is using a stack to store left and right children, and push right child first so that it is processed after the left child.

Java Solution

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
 
public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> returnList = new ArrayList<Integer>();
 
        if(root == null)
            return returnList;
 
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
 
        while(!stack.empty()){
            TreeNode n = stack.pop();
            returnList.add(n.val);
 
            if(n.right != null){
                stack.push(n.right);
            }
            if(n.left != null){
                stack.push(n.left);
            }
 
        }
        return returnList;
    }
}

Category >> Algorithms

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abgalphabet on 2014-3-15

it would be better to use recursion rather than imperative to solve binary tree traversal.
here is my version.

private List traversePreOrder(TreeNode root) {
if(root == null) return Collections.EMPTY_LIST;

List ordered = new ArrayList();

ordered.add(root.value);
ordered.addAll(traversePreOrder(root.left));
ordered.addAll(traversePreOrder(root.right));

return ordered;

}
Prabhat Kumar Kashyap on 2015-1-18

//Recursive Preorder, Inorder, PostOrder traversals.

import java.util.ArrayList;

import java.util.List;

public class TreeTraversal {

static List orderedList = new ArrayList();

public static void preOrderTraversal(TreeNode root){

if(root == null)

return;

orderedList.add(root.val);

preOrderTraversal(root.left);

preOrderTraversal(root.right);

}

public static void inOrderTraversal(TreeNode root){

if(root == null)

return;

inOrderTraversal(root.left);

orderedList.add(root.val);

inOrderTraversal(root.right);

}

public static void postOrderTraversal(TreeNode root){

if(root == null)

return;

postOrderTraversal(root.left);

postOrderTraversal(root.right);

orderedList.add(root.val);

}

public static void main(String[] args) {

TreeNode root = new TreeNode(2);

TreeNode left = new TreeNode(1);

TreeNode right = new TreeNode(3);

root.left = left;

root.right = right;

System.out.print(“nPreorder traversal of 2 1 3: “);

preOrderTraversal(root);

for(Integer i:orderedList)

System.out.print(i + ” “);

orderedList.clear();

System.out.print(“nIneorder traversal of 2 1 3: “);

inOrderTraversal(root);

for(Integer i:orderedList)

System.out.print(i + ” “);

orderedList.clear();

System.out.print(“nPostorder traversal of 2 1 3: “);

postOrderTraversal(root);

for(Integer i:orderedList)

System.out.print(i + ” “);

}
}

/*
OUTPUT:
Preorder traversal of 2 1 3: 2 1 3
Ineorder traversal of 2 1 3: 1 2 3
Postorder traversal of 2 1 3: 1 3 2
*/
fishwen on 2015-9-6

http://www.javabeat.net/binary-search-tree-traversal-java/
Bendells on 2017-2-13

OFC but If you read the problem on leetCode, you would see that they asked to try an iterative solution!

Leetcode – Binary Tree Preorder Traversal (Java)

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