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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?
这道题考察用 stack实现前序遍历,代码如下:
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer>(); Stack<TreeNode> stack =new Stack<TreeNode>(); if(root == null){ return list; } stack.push(root); while(!stack.isEmpty()){ TreeNode node = stack.pop(); list.add(node.val); if(node.right != null){ stack.push(node.right); } if(node.left != null){ stack.push(node.left); } } return list; } }
LeetCode-Binary Tree Preorder Traversal
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原文地址:http://www.cnblogs.com/incrediblechangshuo/p/4331980.html