Binary Tree Preorder Traversal(lintcode 66)

Description

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

Example

Given:

    1
   / \
  2   3
 / \
4   5

return [1,2,4,5,3].

Interface

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
    }
}

Solution

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> result = new ArrayList<Integer>();

        traverse(root, result);
        return result;
    }

    private void traverse(TreeNode root, ArrayList<Integer> result) {
        if (root == null) {
            return;
        }

        if (root.left == null && root.right == null) {
            result.add(root.val);
            return;
        }

        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
}