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https://oj.leetcode.com/problems/binary-tree-preorder-traversal/
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?
解题思路：
很简单，递归。
/** * 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 list = new ArrayList<Integer>(); if(root == null){ return list; } list.add(root.val); list.addAll(preorderTraversal(root.left)); list.addAll(preorderTraversal(root.right)); return list; } }
但原题提出不用递归的方法，即遍历。使用stack，遇到一个节点，首先list.add(root.val)，如果左节点不为空，下移到左节点，同时如果右节点不为空，将右节点推入stack。如果左节点为空，从stack中弹出最上面的节点作为下一个节点，循环处理。直到stack为空，而且当前处理节点为空。
/** * 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 list = new ArrayList<Integer>(); Stack<TreeNode> stack = new Stack<TreeNode>(); while(!stack.empty() || root != null){ if(root == null){ return list; } list.add(root.val); if(root.right != null){ stack.push(root.right); } if(root.left != null){ root = root.left; }else{ if(!stack.empty()){ root = stack.pop(); }else{ root = null; } } } return list; } }
Binary Tree Preorder Traversal
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原文地址：http://www.cnblogs.com/NickyYe/p/4227131.html
