Leetcode Binary Tree Preorder traversal first root traversal

Test instructions: Give a tree the result of its first root traversal.

Ideas:

(1) Deep search method:

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  */Ten classSolution { One  Public: Avector<int> Preordertraversal (treenode*root) { -         if(!root)returnvector<int>(); -vector<int> Ans (1,root->val); thevector<int> Tmp=preordertraversal (root->Left ); - Ans.insert (Ans.end (), Tmp.begin (), Tmp.end ()); -Tmp=preordertraversal (root->Right ); - Ans.insert (Ans.end (), Tmp.begin (), Tmp.end ()); +         returnans; -     } +  A};

AC Code

(2) is still a deep search method:

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  */Ten classSolution { One  Public: Avector<int>ans; -     voidDFS (treenode*t) -     { the         if(!t)return; -Ans.push_back (t->val); -DFS (t->Left ); -DFS (t->Right ); +     } -vector<int> Preordertraversal (treenode*root) { + DFS (root); A         returnans; at     } -};

AC Code

(3) Iterative method:

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  */Ten classSolution { One  Public: Avector<int> Preordertraversal (treenode*root) { -         if(!root)returnvector<int>(); -  thestack<pair<treenode*,int>>Stac; -Stac.push (Make_pair (Root,0));//The second parameter is used to record that it has traversed the left/right child -vector<int>ans; -          while( !stac.empty ()) +         { -treenode* cur=Stac.top (). First; +             if(Stac.top (). second==0) Ans.push_back (cur->val); A  at             if(Cur->left && stac.top (). second==0)//didn't walk through the kids. -             { -Cur=cur->Left ; -Stac.top (). second=1; -Stac.push (Make_pair (cur,0)); -             } in             Else if(Cur->right && stac.top () .second<2)//walk through the left child or have no left child to come in -             { toCur=cur->Right ; +Stac.top (). second=2; -Stac.push (Make_pair (cur,0)); the             } *             ElseStac.pop ();//None of the above two can get in or go through or have no children . $         }Panax Notoginseng         returnans; -     } the};

AC Code

(4) More iterative method:

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  */Ten classSolution { One  Public: Avector<int> Preordertraversal (treenode*root) { -         if(!root)returnvector<int>(); -  thevector<int>ans; -Stack<treenode *>Stac; - Stac.push (root); -  +          while( !stac.empty ()) -         { +TreeNode *t=stac.top (); AAns.push_back (t->val); atStac.pop ();//All I need is a child to push the stack, father useless -             if(t->right) Stac.push (t->Right ); -             if(t->left) Stac.push (t->Left ); -         } -         returnans; -     } in};

AC Code

Leetcode Binary Tree Preorder traversal first root traversal