[LeetCode-JAVA] Binary Tree Inorder Traversal

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/