标签：des style class blog code color
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:
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 Stack<TreeNode> store = new Stack<TreeNode>(); 14 if (root == null) return result; 15 store.push(root); 16 while (!store.empty()) { 17 TreeNode temp = store.pop(); 18 result.add(temp.val); 19 if (temp.right != null) store.push(temp.right); 20 if (temp.left != null) store.push(temp.left); 21 } 22 return result; 23 } 24 }
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 }
Leetcode: Binary Tree Preorder Traversal,布布扣,bubuko.com
Leetcode: Binary Tree Preorder Traversal
标签：des style class blog code color
原文地址：http://www.cnblogs.com/EdwardLiu/p/3796289.html
