LeetCode—Binary Tree Zigzag Level Order Traversal 二叉树的Z扫描

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]
]

题目还是比较简单，主要就是一个层次扫描然后进行翻转的过程

这里有点类似层序扫描，在层序扫描中主要是利用了一个queue进行存储节点，然后读出对应的val，这里需要返回的类型需要具体按层给出，所以就需要用到一个vector将每一层的节点进行记录

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> res;
    int flag= 0;
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        res.clear();
        if(root == NULL)
        {
            return res;
        }
        vector<TreeNode*>pNode;
        pNode.push_back(root);
        vector<int> tmp;
        tmp.push_back(root->val);
        res.push_back(tmp);
        flag++;
        getNode(pNode);
        return res;
    }
    void getNode(vector<TreeNode*> &pNode)
    {
        if(pNode.empty())
        {
            return;
        }
        vector<TreeNode*> Node;
        vector<TreeNode*>::iterator it = pNode.begin();
        vector<int> tmp;
        for(;it != pNode.end();it++)
        {
            TreeNode * root = *it;
            if(root->left)
            {
                tmp.push_back(root->left->val);
                Node.push_back(root->left);
            }
            if(root->right)
            {
                tmp.push_back(root->right->val);
                Node.push_back(root->right);
            }
        }
        if(tmp.empty())
        {
            return; //<没有孩子节点，可以返回
        }
        if(flag&1)
        {
            reverse(tmp.begin(),tmp.end());
        }
        flag++;
        res.push_back(tmp);
        getNode(Node);
    }
};

主要是利用了一个vector的翻转操作

还看到一个用queue进行操作的方法，这里是在每一层操作的最后面放上一个NULL节点，用于判断这一层的数据解完了

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int> > result;
        vector<int> temp;
        result.clear();
        if(root==NULL)
            return result;
        queue<TreeNode *> pTemp;
        int flag=0;
        pTemp.push(root);
        pTemp.push(NULL);
        while(!pTemp.empty())
        {
            TreeNode *curNode=pTemp.front();
            pTemp.pop();
            if(curNode!=NULL)
            {
                temp.push_back(curNode->val);
                if(curNode->left)
                    pTemp.push(curNode->left);
                if(curNode->right)
                    pTemp.push(curNode->right);
            }
            else
            {
                if(!temp.empty())
                {
                    pTemp.push(NULL);
                    if(flag==1)
                    {
                        reverse(temp.begin(),temp.end());
                    }
                    result.push_back(temp);
                    flag=1-flag;
                    temp.clear();
                }
            }
        }
        return result;
    }
};