Leetcode populating Next right pointers in each Node (tips)

Test instructions

To a tree full of two, requires that each layer of the node from left to right with the next pointer, the end of the layer pointing to null.

Ideas:

can be recursive or iterative. It is necessary to observe that Next's left child happens to be next to the right child of this node.

(1) Recursion: This is faster.

1 /**2 * Definition for binary tree with next pointer.3 * struct Treelinknode {4 * int val;5 * Treelinknode *left, *right, *next;6 * Treelinknode (int x): Val (x), left (null), Right (null), Next (null) {}7  * };8  */9 classSolution {Ten  Public: One     voidDFS (treelinknode* root,treelinknode*t) A     { -         if(Root->next==null && t!=null) root->next=t->Left ; -         if(root->Left ) the         { -Root->left->next=root->Right ; -DFS (root->Left , NULL); -DFS (root->right,root->next); +         } -     } +     voidConnect (Treelinknode *root) { A         if(Root) DFS (root,null); at     } -};

AC Code

(2) Recursion: this brevity.

1 /**2 * Definition for binary tree with next pointer.3 * struct Treelinknode {4 * int val;5 * Treelinknode *left, *right, *next;6 * Treelinknode (int x): Val (x), left (null), Right (null), Next (null) {}7  * };8  */9 classSolution {Ten  Public: One     voidConnect (Treelinknode *root) { A         if(Root && root->Left ) -         { -Root->left->next=root->Right ; the             if(root->next) -Root->right->next=root->next->Left ; -Connect (root->Left ); -Connect (root->Right ); +         } -  +          A     } at};

AC Code

(3) Iteration: This is more hanging.

　　

Leetcode populating Next right pointers in each Node (tips)