094_BinaryTreeInorderTraversal94.java

/**
 * Given a binary tree, return the inorder traversal of its nodes' values.
 *
 * For example:
 * Given binary tree [1,null,2,3],
 *    1
 *     \
 *      2
 *     /
 *    3
 * return [1,3,2].
 *
 * Note: Recursive solution is trivial, could you do it iteratively?
 */


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */


import java.util.ArrayList;
import java.util.List;
import java.util.Stack;


public class BinaryTreeInorderTraversal94 {
    public List<Integer> inorderTraversal(TreeNode root) {
        if (node == null) {
            return new ArrayList<Integer>();
        }

        List<Integer> result =  new ArrayList<>();

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

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

        return result;
    }

    /**
     * https://discuss.leetcode.com/topic/6478/iterative-solution-in-java-simple-and-readable
     */
    public List<Integer> inorderTraversal2(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;

        while (cur != null || !stack.empty()){
            while (cur != null){
                stack.add(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            list.add(cur.val);
            cur = cur.right;
        }

        return list;
    }


    public List<Integer> inorderTraversal3(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        inorderTraversal(root, res);
        return res;
    }

    private void inorderTraversal(TreeNode root, List<Integer> res) {
        if (root == null) return;
        inorderTraversal(root.left, res);
        res.add(root.val);
        inorderTraversal(root.right, res);
    }

}