标签：
https://oj.leetcode.com/problems/binary-tree-inorder-traversal/
Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
解题思路：
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> inorderTraversal(TreeNode root) { List list = new ArrayList<Integer>(); if(root == null){ return list; } list.addAll(inorderTraversal(root.left)); list.add(root.val); list.addAll(inorderTraversal(root.right)); return list; } }
非递归方法
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<Integer> (); Stack<TreeNode> stack = new Stack<TreeNode> (); while(!stack.empty() || root != null){ while(root != null){ stack.push(root); root = root.left; } if(!stack.empty()){ root = stack.pop(); list.add(root.val); root = root.right; } } return list; } }
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原文地址：http://www.cnblogs.com/NickyYe/p/4227226.html
