BinaryTreeInorderTraversal.java

/*
 Author:     King, wangjingui@outlook.com
 Date:       Dec 20, 2014
 Problem:    Binary Tree Inorder Traversal
 Difficulty: Easy
 Source:     https://oj.leetcode.com/problems/binary-tree-inorder-traversal/
 Notes:
 Given a binary tree, return the inorder traversal of its nodes' values.
 For example:
 Given binary tree {1,#,2,3},
 1
  \
   2
  /
 3
 return [1,3,2].
 Note: Recursive solution is trivial, could you do it iteratively?

 Solution: 1. Recursive solution.      Time: O(n), Space: O(n).
           2. Iterative way (stack).   Time: O(n), Space: O(n).
           3. Threaded tree (Morris).  Time: O(n), Space: O(1).
 */

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal_1(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) return res;
        inorder(root, res);
        return res;
    }
    public void inorder(TreeNode root, List<Integer> res) {
        if (root == null) return;
        inorder(root.left, res);
        res.add(root.val);
        inorder(root.right, res);
    }
    public List<Integer> inorderTraversal_2(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stk = new Stack<TreeNode>();
        TreeNode cur = root;
        while (stk.isEmpty() == false || cur != null) {
            if (cur != null) {
                stk.push(cur);
                cur = cur.left;
            } else {
                cur = stk.pop();
                res.add(cur.val);
                cur = cur.right;
            }
        }
        return res;
    }
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        TreeNode cur = root;
        while (cur != null) {
            if (cur.left == null) {
                res.add(cur.val);
                cur = cur.right;
            } else {
                TreeNode node = cur.left;
                while (node.right != null && node.right != cur)
                    node  = node.right;
                if (node.right == null) {
                    node.right = cur;
                    cur = cur.left;
                } else {
                    res.add(cur.val);
                    node.right = null;
                    cur = cur.right;
                }
            }
        }
        return res;
    }
}