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题目：
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].
思路：题中规定用循环代替递归求解，用辅助stack来存储遍历到的节点，如果不为空则入栈，否则出栈赋值，并指向其右节点。
代码：
public class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> req = new ArrayList<Integer>(); if(root == null) return req; Stack<TreeNode> stack = new Stack<TreeNode>(); while(!stack.isEmpty() || root != null){ if(root != null){ stack.push(root); root = root.left; }else{ TreeNode temp = stack.pop(); req.add(temp.val); root = temp.right; //最重要的一步 } } return req; } }
参考链接： http://www.programcreek.com/2012/12/leetcode-solution-of-binary-tree-inorder-traversal-in-java/
[LeetCode-JAVA] Binary Tree Inorder Traversal
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原文地址：http://www.cnblogs.com/TinyBobo/p/4518975.html
