题意：

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?

    思路：

    二叉树的非递归中序遍历，和前序遍历很相似，把访问值的位置改了就好。

    代码：

/**
 * 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> ans;
        stack<TreeNode *> s;
        TreeNode *p = root;
        while(p != NULL || !s.empty())
        {
            while(p != NULL)
            {
                s.push(p);
                p = p->left;
            }
            if(!s.empty())
            {
                p = s.top();
                s.pop();
                ans.push_back(p->val);
                p = p->right;
            }
        }
        return ans;
    }
};

      Python：（用 Python 写栈好神奇）

# 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):
        ans = []
        s = []
        while root != None or s:
            while root != None:
                s.append(root)
                root = root.left
            if s:
                root = s[-1]
                s.pop()
                ans.append(root.val)
                root = root.right
        return ans

【LeetCode】Binary Tree Inorder Traversal