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 = 0;
        queue<Node>q;
        q.push(Node(root, 0));
        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 + 1));
            if(tmpNode.node->right)
                q.push(Node(tmpNode.node->right, tmpNode.level + 1));
            if(dep < tmpNode.level){
                if(dep % 2){
                    reverse(tmp.begin(), tmp.end());
                }
                ret.push_back(tmp);
                tmp.clear();
                dep = tmpNode.level;
            }
            tmp.push_back(tmpNode.node->val);
            q.pop();
        }
        if(dep % 2){
            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);
    }
}