LeetCode Binary Tree Preorder Traversal 先根遍历

 

 

 

题意：给一棵树，求其先根遍历的结果。

思路：

（1）深搜法：

 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> preorderTraversal(TreeNode* root) {

13         if(!root)  return vector<int>();

14         vector<int> ans(1,root->val);

15         vector<int> tmp=preorderTraversal(root->left);

16         ans.insert(ans.end(),tmp.begin(),tmp.end());

17         tmp=preorderTraversal(root->right);

18         ans.insert(ans.end(),tmp.begin(),tmp.end());

19         return ans;

20     }

21 

22 };

AC代码

 

（2）依然是深搜法：

 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> ans;

13     void DFS(TreeNode* t)

14     {

15         if(!t)  return;

16         ans.push_back(t->val);

17         DFS(t->left);

18         DFS(t->right);

19     }

20     vector<int> preorderTraversal(TreeNode* root) {

21         DFS(root);

22         return ans;

23     }

24 };

AC代码

 

（3）迭代法：

 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> preorderTraversal(TreeNode* root) {

13         if(!root)  return vector<int>();

14 

15         stack<pair<TreeNode*,int>>  stac;

16         stac.push(make_pair(root,0));//第二个参数用于记录其已经遍历了左/右孩子

17         vector<int> ans;

18         while( !stac.empty() )

19         {

20             TreeNode* cur=stac.top().first;

21             if( stac.top().second==0 )  ans.push_back(cur->val);

22 

23             if(cur->left && stac.top().second==0)//没有遍历过孩子的进来

24             {

25                 cur=cur->left;

26                 stac.top().second=1;

27                 stac.push(make_pair(cur,0));

28             }

29             else if(cur->right && stac.top().second<2)//遍历过左孩子或者没有左孩子的才进来

30             {

31                 cur=cur->right;

32                 stac.top().second=2;

33                 stac.push(make_pair(cur,0));

34             }

35             else    stac.pop();//以上两个都进不去，要么遍历完，要么没有孩子

36         }

37         return ans;

38     }

39 };

AC代码

 

（4）更叼的迭代法：

 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> preorderTraversal(TreeNode* root) {

13         if(!root)  return vector<int>();

14 

15         vector<int> ans;

16         stack<TreeNode *>  stac;

17         stac.push(root);

18 

19         while( !stac.empty() )

20         {

21             TreeNode *t=stac.top();

22             ans.push_back(t->val);

23             stac.pop();//只需要孩子都压栈，父亲无用

24             if(t->right)    stac.push(t->right);

25             if(t->left)     stac.push(t->left);

26         }

27         return ans;

28     }

29 };

AC代码