Binary Tree Inorder/Preorder/Postorder Traversal

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?

Recursion:

public class Solution {
    //Time: O(n)  Space: O(depth of tree)
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        inorder(root, res);
        return res;
    }
    
    private void inorder(TreeNode root, ArrayList<Integer> res) {
        if (root == null) {
            return;
        }
        //res.add(root.val);  pre
        inorder(root.left, res);
        //res.add(root.val);  in
        inorder(root.right, res);
        //res.add(root.val);  post
    }
}

Iterative: Inorder
使用一个stack，如果有左子树，就把当前点放进stack，反之pop出一个node，检查它的右子树。

public class Solution {
    //Time: O(n)  Space: O(depth of tree)
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while (!stack.isEmpty() || p != null) {
            if (p != null) {
                stack.push(p);
                p = p.left;
            } else {
                p = stack.pop();
                res.add(p.val);
                p = p.right;
            }
        }
        return res;
    }
}

Iterative: Preorder
用一个stack，当前node被pop出后，先push右子树节点，再push左子树节点

public class Solution {
    //Time: O(n)  Space: O(n)
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            if (cur.right != null) { //push right first
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
            res.add(cur.val);
        }
        return res;
    }
}

Iterative: Postorder

public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode pre = null;
        TreeNode cur = root;
        stack.push(root);
        while (!stack.isEmpty()) {
            cur = stack.peek();
            if (pre == null || pre.left == cur || pre.right == cur) {
                if (cur.left != null) {
                    stack.push(cur.left);
                } else if (cur.right != null) {
                    stack.push(cur.right);
                }
            } else if (cur.left == pre) {
                if (cur.right != null) {
                    stack.push(cur.right);
                }
            } else {
                res.add(cur.val);
                stack.pop();
            }
            pre = cur;
        }
        return res;
    }
}