Leetcode 103. Binary Tree Zigzag level Order traversal

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>> Zigzaglevelorder (treenode*root) { -vector<int>Row; -vector<vector<int>>v; theQueue<treenode *>Q; -         if(Root = =NULL) -             returnv; - Q.push (root); +TreeNode *temp; -         intLev =0; +          while(!Q.empty ()) { A             intSize =q.size (); at              while(size--) { -temp =Q.front (); - Q.pop (); -Row.push_back (temp->val); -                 if(Temp->left! =NULL) { -Q.push (temp->Left ); in                 } -                 if(Temp->right! =NULL) { toQ.push (temp->Right ); +                 } -             } the             if(Lev%2) { *                 intn =row.size (); $                  for(inti =0; I < n/2; i++) {Panax NotoginsengSwap (Row[i], row[n-i-1]); -                 } the             } + v.push_back (row); Alev++; the row.clear (); +         } -         returnv; $     } $};

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Given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).

For Example:given binary Tree {3,9,20,#,#,15,7} ,

    3   /   9    /   7

Return its zigzag level order traversal as:

[  3],  [20,9],  [15,7]]

Analysis: and sequence traversal the same code, only need to add a few lines of code on the line ~ ~ ~ because the font to store this binary tree ~ ~ ~
So just when the number of rows is even, you need to do all the elements of the current row upside down ~ can be used stack can also be used to exchange the array 22 ~ ~ ~
You just need to invert the row array before you deposit the two-dimensional array vector<vector> V, and then push_back into v.

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>> Zigzaglevelorder (treenode*root) { -vector<int>Row; -vector<vector<int>>v; theQueue<treenode *>Q; -         if(Root = =NULL) -             returnv; - Q.push (root); +TreeNode *temp; -         intLev =0; +          while(!Q.empty ()) { A             intSize =q.size (); at              while(size--) { -temp =Q.front (); - Q.pop (); -Row.push_back (temp->val); -                 if(Temp->left! =NULL) { -Q.push (temp->Left ); in                 } -                 if(Temp->right! =NULL) { toQ.push (temp->Right ); +                 } -             } the             if(Lev%2) { *                 intn =row.size (); $                  for(inti =0; I < n/2; i++) {Panax NotoginsengSwap (Row[i], row[n-i-1]); -                 } the             } + v.push_back (row); Alev++; the row.clear (); +         } -         returnv; $     } $};

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Leetcode 103. Binary Tree Zigzag level Order traversal