递归版本

1#LeetCode 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.
Note: Recursive solution is trivial, could you do it iteratively?

Preorder: root, left, right

Recursive version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        traverse (root, result);
        return result;
    }
    public void traverse (TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
}

Non-recursive version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            result.add(node.val);
            // 关键点：要先压入右孩子，再压入左孩子，这样在出栈时会先打印左孩子再打印右孩子 
            if (node.right != null) {
                stack.push(node.right);
            }
            if (node.left != null) {
                stack.push(node.left);
            }
        }
        return result;
    }
}

2#LeetCode 94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.
Note: Recursive solution is trivial, could you do it iteratively?

Inorder: left, root, right

Recursive version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        traverse (root, result);
        return result;
    }
    public void traverse (TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        traverse(root.left, result);
        result.add(root.val);
        traverse(root.right, result);
    }
}

用栈先把根节点的所有左孩子都添加到栈内， 
然后输出栈顶元素，再处理栈顶元素的右子树 
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        if (root == null) {
            return result;
        }
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.add(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            result.add(cur.val);
            cur = cur.right;
        }
        return result;
    }
}

3#145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.
Note: Recursive solution is trivial, could you do it iteratively?

postorder: left, right, root

Recursive version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        traverse(root, result);
        return result;
    }
    public void traverse (TreeNode root, List<Integer> result) {
        if (root == null) {
            return;
        }
        traverse(root.left, result);
        traverse(root.right, result);
        result.add(root.val);
    }
}