Leetcode 102 binary Tree level Order traversal two forks +bfs

Hierarchical traversal of binary tree

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<vector<int>> Levelorder (treenode*root) { -vector<vector<int>>v; -         if(!root)returnv; thevector<int>T; -          -T.push_back (root->val); - V.push_back (t); +Root->val =1; -          +Queue<treenode*>Q; A Q.push (root); at          -          while(!Q.empty ()) { -treenode* now =Q.front (); - Q.pop (); -             if(!now)Continue; -              inQ.push (now->Left ); -Q.push (now->Right ); to             if(Now->val <v.size ()) { +                 if(now->left) V[now->val].push_back (now->left->val); -                 if(now->right) V[now->val].push_back (now->right->val); the             } *             Else{ $vector<int>T; Panax Notoginseng                 if(now->left) T.push_back (now->left->val); -                 if(now->right) T.push_back (now->right->val); the                 if(T.size ()! =0) V.push_back (t); +             } A             if(now->left) Now->left->val = Now->val +1; the             if(now->right) Now->right->val = Now->val +1; +         } -         //Reverse (V.begin (), V.end ()); $         returnv; $     } -};

Leetcode 102 binary Tree level Order traversal two forks +bfs