[LeetCode] Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

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

   1
    \
     2
    /
   3

 

return [3,2,1].

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

思路： 用r表示最近访问过的节点。

　　　时间复杂度O(n)， 空间复杂度O(n)

 1 /**
 2  * Definition for binary tree
 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> postorderTraversal(TreeNode *root) {
13         vector<int> result;
14         stack<TreeNode *> s;
15         TreeNode *p = root, *r = NULL;
16         
17         while (p != NULL || !s.empty()) {
18             if (p!= NULL) {
19                 s.push(p);
20                 p = p->left;
21             } else {
22                 p = s.top();
23                 if (p->right != NULL && p->right != r) {
24                     p = p->right;
25                 } else {
26                     result.push_back(p->val);
27                     s.pop();
28                     r = p;
29                     p = NULL;
30                 }
31             }
32         }
33         
34         return result;
35     }
36 };

 

posted @ 2015-01-16 19:53  vincently  阅读(238)  评论(0)  编辑  收藏  举报