同num8一样，此题考查的是二叉树的中序遍历，即先左子树再节点再右子树、

使用迭代法时，采用将节点和左子树均压入栈的方法，当左子树为NULL时，将top节点弹出，并存入结果列表，将next指针指向该节点的右子树

代码如下：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> rs;                  #存储结果
        stack<TreeNode *> nstack;
        TreeNode *next=root;
        while(next||!nstack.empty()){
            while(next){
                nstack.push(next);
                next=next->left;
            }
            next=nstack.top();
            nstack.pop();
            rs.push_back(next->val);#结果rs中先存入左子树节点值
            next=next->right;
        }
        return rs;
    }
};

递归法就比较简单了，代码如下：

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of integers
    def inorderTraversal(self, root):
        rs=[]
        if root!=None:
            rs=self.inorderTraversal(root.left)+[root.val]+self.inorderTraversal(root.right)
        return rs

Leetcode_num9_Binary Tree Inorder Traversal