Binary Tree Zigzag Level Order Traversal

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},

  1. 3
  2. / \
  3. 9 20
  4. / \
  5. 15 7

return its zigzag level order traversal as:

  1. [
  2. [3],
  3. [20,9],
  4. [15,7]
  5. ]
 
 
利用队列,进行层次遍历
根据层的奇偶性,判断是否需要翻转
 
 
  1. /**
  2. * Definition for binary tree
  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<vector<int> > zigzagLevelOrder(TreeNode *root) {
  13.  
  14. vector<vector<int> > result;
  15. queue<TreeNode*> q;
  16.  
  17. if(root==NULL) return result;
  18. q.push(root);
  19.  
  20. int numNode=;
  21.  
  22. bool isOdd=true;
  23.  
  24. while(!q.empty())
  25. {
  26. vector<int> cur;
  27. int nextLevelCount=;
  28. for(int i=;i<numNode;i++)
  29. {
  30. TreeNode *node=q.front();
  31. cur.push_back(node->val);
  32. q.pop();
  33. if(node->left!=NULL)
  34. {
  35. q.push(node->left);
  36. nextLevelCount++;
  37. }
  38.  
  39. if(node->right!=NULL)
  40. {
  41. q.push(node->right);
  42. nextLevelCount++;
  43. }
  44. }
  45.  
  46. if(isOdd)
  47. {
  48. result.push_back(cur);
  49. }
  50. else
  51. {
  52. reverse(cur.begin(),cur.end());
  53. result.push_back(cur);
  54. }
  55.  
  56. isOdd=!isOdd;
  57.  
  58. numNode=nextLevelCount;
  59. }
  60.  
  61. return result;
  62.  
  63. }
  64. };

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