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Given a binary tree, return the preorder traversal of its nodes‘ values.
Given:
1
/ 2 3
/ 4 5
return [1,2,4,5,3]
.
Can you do it without recursion?
1. recursive
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: Preorder in ArrayList which contains node values. */ public ArrayList<Integer> preorderTraversal(TreeNode root) { // write your code here ArrayList<Integer> result = new ArrayList<Integer>(); if(root == null) return result; helper(root, result); return result; } public void helper(TreeNode root, ArrayList<Integer> result){ if(root == null) return; else{ result.add(root.val); helper(root.left, result); helper(root.right, result); return; } } }
2. iterative
方法和中序遍历差不多
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: Preorder in ArrayList which contains node values. */ public ArrayList<Integer> preorderTraversal(TreeNode root) { // write your code here ArrayList<Integer> result = new ArrayList<Integer>(); if(root == null) return result; TreeNode p = null; Stack<TreeNode> stack = new Stack<TreeNode>(); stack.push(root); while(p != null || !stack.isEmpty()){ if(p != null){ result.add(p.val); if(p.right != null){ stack.push(p.right); } p = p.left; } else{ p = stack.pop(); } } return result; } }
lintcode-easy-Binary Tree Preorder Traversal
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原文地址:http://www.cnblogs.com/goblinengineer/p/5205302.html