LeetCode: Binary Tree Inorder Traversal

一次过

 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     void dfs(TreeNode *root, vector<int> &ret) {
13         if (!root) return;
14         dfs(root->left, ret);
15         ret.push_back(root->val);
16         dfs(root->right, ret);
17     }
18     vector<int> inorderTraversal(TreeNode *root) {
19         // Start typing your C/C++ solution below
20         // DO NOT write int main() function
21         vector<int> ret;
22         dfs(root, ret);
23         return ret;
24     }
25 };

写了一个不需要递归的代码，这段代码更好

 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> inorderTraversal(TreeNode *root) {
13         // Start typing your C/C++ solution below
14         // DO NOT write int main() function
15         stack<TreeNode *> S;
16         vector<int> ret;
17         if (!root) return ret;
18         S.push(root);
19         while (!S.empty()) {
20             TreeNode *tmp = S.top();
21             if (tmp->left) {
22                 S.push(tmp->left);
23                 tmp->left = NULL;
24             }
25             else {
26                 ret.push_back(tmp->val);
27                 S.pop();
28                 if (tmp->right) S.push(tmp->right);
29             }
30         }
31         return ret;
32     }
33 };

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<int> InorderTraversal(TreeNode root) {
12         Stack<TreeNode> S = new Stack<TreeNode>();
13         List<int> ans = new List<int>();
14         if (root == null) return ans;
15         S.Push(root);
16         while(S.Count != 0)
17         {
18             TreeNode top = S.Peek();
19             if (top.left != null) {
20                 S.Push(top.left);
21                 top.left = null;
22             }
23             else {
24                 ans.Add(top.val);
25                 S.Pop();
26                 if (top.right != null) S.Push(top.right);
27             }
28         }
29         return ans;
30     }
31 }

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posted @ 2013-03-18 09:40 ying_vincent 阅读(157) 评论(0) 编辑收藏举报

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