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?
Analysis: 第一反应肯定是recursion(递归), 非常直接,但是题目要求不能用递归。如果要使用迭代的方法来写preorder traversal,最大的问题是要如何确定遍历节点的顺序,因为树的pre-order traversal其实很类似图的DFS,DFS可以用Stack来写,所以这里写pre-order traversal也可以用stack来实现迭代的写法。
Iterative: 其实就是用一个栈来模拟递归的过程。所以算法时间复杂度也是O(n),空间复杂度是栈的大小O(logn)。
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<Integer> preorderTraversal(TreeNode root) { 12 List<Integer> res = new ArrayList<>(); 13 TreeNode p = root; 14 Stack<TreeNode> stack = new Stack<>(); 15 while (p!=null || !stack.isEmpty()) { 16 if (p != null) { 17 stack.push(p); 18 res.add(p.val); 19 p = p.left; 20 } 21 else { 22 TreeNode node = stack.pop(); 23 p = node.right; 24 } 25 } 26 return res; 27 } 28 }
recursion:
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public ArrayList<Integer> preorderTraversal(TreeNode root) { 12 ArrayList<Integer> result = new ArrayList<Integer>(); 13 preorder(root, result); 14 return result; 15 } 16 17 public void preorder(TreeNode root, ArrayList<Integer> result) { 18 if (root == null) return; 19 result.add(root.val); 20 preorder(root.left, result); 21 preorder(root.right, result); 22 } 23 }