LeetCode OJ：Binary Tree Zigzag Level Order Traversal（折叠二叉树遍历）

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  20
    /  \
   15   7

return its zigzag level order traversal as:

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

简单的bfs而已，不过在bfs的过程中应该注意将相应的数组reverse一下，其实都到最终结果之后在隔行reverse也是可以的，下面给出非递归的版本，用递归同样也很好实现：

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
     struct Node
     {
         TreeNode * node;
         int level;
         Node(){}
         Node(TreeNode * n, int lv)
         : node(n), level(lv){}
     };
 public:
     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
         vector<vector<int>> ret;
         if(!root) return ret;
         vector<int> tmp;
         int dep = ;
         queue<Node>q;
         q.push(Node(root, ));
         while(!q.empty()){
             Node tmpNode = q.front(); //非递归的使用bfs，借助队列特性
             if(tmpNode.node->left)
                 q.push(Node(tmpNode.node->left, tmpNode.level + ));
             if(tmpNode.node->right)
                 q.push(Node(tmpNode.node->right, tmpNode.level + ));
             if(dep < tmpNode.level){
                 if(dep % ){
                     reverse(tmp.begin(), tmp.end());
                 }
                 ret.push_back(tmp);
                 tmp.clear();
                 dep = tmpNode.level;
             }
             tmp.push_back(tmpNode.node->val);
             q.pop();
         }
         if(dep % ){
             reverse(tmp.begin(), tmp.end());
         }
         ret.push_back(tmp);
         return ret;
     }
 };

java版本的代码如下所示，这里用的是dfs而非bfs，在最后重新reverse就可以了：

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         dfs(ret, 1, root);
         for(int i = 0; i < ret.size(); ++i){
             if(i%2 != 0) Collections.reverse(ret.get(i));
         }
         return ret;
     }

     void dfs(List<List<Integer>>ret, int dep, TreeNode root){
         if(root == null)
             return;
         if(ret.size() < dep){
             List<Integer> list = new ArrayList<Integer>();
             list.add(root.val);
             ret.add(list);
         }else
             ret.get(dep - 1).add(root.val);
         dfs(ret, dep + 1, root.left);
         dfs(ret, dep + 1, root.right);
     }
 }