[LeetCode]Binary Tree Level Order Traversal (I II)和Binary Tree Zigzag Level Order Traversal

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Binary Tree Level Order Traversal题目描述：

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).

解释：
针对二叉树的层次遍历，一般采用队列数据结构，首先根节点被压入队列，当根节点从队首弹出时，该根节点的左右节点分别从队尾压入，利用这一原理实现层次遍历，代码如下：

C++代码：

/**
 * 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 {
public:
    vector<vector<int>> levelOrder(TreeNode* root) 
    {
        vector<vector<int>>result;
        vector<int>temp;
        queue<TreeNode*>que;
        if(root==NULL)
        {
            return result;
        }
        que.push(root);
        while(!que.empty())
        {
            int size= que.size();
            for(int i = 0;i<size;i++)
            {
                TreeNode*node = que.front();
                que.pop();
                temp.push_back(node->val);  
                if(node->left!=NULL)
                {
                    que.push(node->left);
                }
                if(node->right!=NULL)
                {
                    que.push(node->right);
                }  
            }
            result.push_back(temp);
            temp.clear();
        }
        return result;
    }
};

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Binary Tree Level Order Traversal ||题目描述：

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

解释：
从例子中可以看出，这道题目是上面题目层次遍历的升级版，可以先按照上面题目的思路求出结果，再将结果翻转下就是该题的结果。

C++代码：

/**
 * 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 {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) 
    {
        vector<vector<int>>result;
        vector<int>temp;
        queue<TreeNode*>que;
        if(root==NULL)
        {
            return result;
        }
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            for(int i = 0 ;i<size;i++)
            {
                TreeNode *node = que.front();
                que.pop();
                temp.push_back(node->val);
                if(node->left)
                {
                    que.push(node->left);
                }
                if(node->right)
                {
                    que.push(node->right);
                }
            }
            result.push_back(temp);
            temp.clear();
        }
        reverse(result.begin(),result.end());
        return result;
    }
};

运行结果：

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Binary Tree Zigzag Level Order Traversal题目描述：

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,null,null,15,7],

C++代码：

/**
 * 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 {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) 
    {
        vector<vector<int>>result;
        vector<int>temp;
        queue<TreeNode*>que;
        if(root==NULL)
        {
            return result;
        }
        que.push(root);
        while(!que.empty())
        {
            int size = que.size();
            for(int i = 0;i<size;i++)
            {
                TreeNode *node = que.front();
                que.pop();
                temp.push_back(node->val);
                if(node->left)
                {
                    que.push(node->left);
                }
                if(node->right)
                {
                    que.push(node->right);
                }
            }
            result.push_back(temp);
            temp.clear();
        }
        for(int i = 0;i<result.size();i++)
        {
            if(i%2==1)
            {
               reverse(result[i].begin(),result[i].end());
            }
        }
        return result;  
    }
};

运行结果：