LeetCode94:Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

二叉树的中序遍历非递归实现，悲催的是只知道需要使用栈来实现，但是具体的代码如何编写却忘记了。函数调用是通过栈来实现的，所以之前使用递归来实现的中序遍历可以转用非递归来实现。

具体的分析可以参考这篇博客。

runtime:0ms

/**
 * Definition for a binary tree node.
 * 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> res;
        if(root==NULL)  
            return res;

        TreeNode *node=root;
        stack<TreeNode*> s;
        while(!s.empty()||node)
        {
            while(node)
            {
                s.push(node);
                node=node->left;
            }

            node=s.top();
            s.pop();
            res.push_back(node->val);
            node=node->right;
        }
        return res;
    }
};