【Leetcode】【Medium】Binary Tree Inorder Traversal

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?

解题思路：

二叉树中序非递归遍历，使用栈来保存遍历到的但是还没有访问其右子树元素的结点；

代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> inorderTraversal(TreeNode* root) {
13         vector<int> vals;
14         stack<TreeNode*> nodes;
15         TreeNode* node = root;
16
17         while (node != NULL || !nodes.empty()) {
18             while (node) {
19                 nodes.push(node);
20                 node = node->left;
21             }
22
23             node = nodes.top();
24             nodes.pop();
25             vals.push_back(node->val);
26             node = node->right;
27         }
28
29         return vals;
30     }
31 };