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},

return [1,2,3].

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

[解题思路]

递归解法

 1 public ArrayList<Integer> preorderTraversal(TreeNode root) {
 2         // IMPORTANT: Please reset any member data you declared, as
 3         // the same Solution instance will be reused for each test case.
 4         ArrayList<Integer> result = new ArrayList<Integer>();
 5         
 6         preorder(root, result);
 7         
 8         return result;
 9     }
10     
11     public void preorder(TreeNode root, ArrayList<Integer> result){
12         if(root != null){
13             result.add(root.val);
14             preorder(root.left, result);
15             preorder(root.right, result);
16         }
17     }

[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

 1 public ArrayList<Integer> preorderTraversal(TreeNode root) {
 2         // IMPORTANT: Please reset any member data you declared, as
 3         // the same Solution instance will be reused for each test case.
 4         ArrayList<Integer> result = new ArrayList<Integer>();
 5         if(root == null){
 6             return result;
 7         }
 8         
 9         Stack<TreeNode> stack = new Stack<TreeNode>();
10         TreeNode cur = root;
11         while(cur != null || !stack.empty()){
12             
13             while(cur != null){
14                 result.add(cur.val);
15                 stack.push(cur);
16                 cur = cur.left;
17             }
18             
19             if(!stack.empty()){
20                 cur = stack.pop();
21                 // if(cur.right != null){
22                     cur = cur.right;
23                 // }
24             }
25         }
26         
27         return result;
28     }