leetcode -- Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

[解题思路]

递归解法

public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ArrayList<Integer> result = new ArrayList<Integer>();
        
        preorder(root, result);
        
        return result;
    }
    
    public void preorder(TreeNode root, ArrayList<Integer> result){
        if(root != null){
            result.add(root.val);
            preorder(root.left, result);
            preorder(root.right, result);
        }
    }

[非递归解法]

1)访问节点cur，并将节点cur入栈；

2)判断节点cur的左孩子是否为空，若为空，则取栈顶节点并进行出栈操作，并将栈顶节点的右孩子置为当前节点cur，循环至1);若不为空，则将cur的左孩子置为当前节点cur；

3)知道cur为null并且栈为空，则遍历结束

public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root == null){
            return result;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while(cur != null || !stack.empty()){
            
            while(cur != null){
                result.add(cur.val);
                stack.push(cur);
                cur = cur.left;
            }
            
            if(!stack.empty()){
                cur = stack.pop();
                // if(cur.right != null){
                    cur = cur.right;
                // }
            }
        }
        
        return result;
    }