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},

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);

     }

 }

巴特西

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

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