Binary Tree Inorder Traversal(lintcode 67)

Description

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

Example

Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

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: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(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: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (root == null) {
            return result;
        }

        ArrayList<Integer> left = inorderTraversal(root.left);
        ArrayList<Integer> right = inorderTraversal(root.right);

        result.addAll(left);
        result.add(root.val);
        result.addAll(right);

        return result;
    }
}