103. Binary Tree Zigzag Level Order Traversal

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每日一题 2019 - 04 - 23

题目：

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],

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2
3
4
5
  3
 / \
9  20
  /  \
 15   7

return its zigzag level order traversal as:

1
2
3
4
5
[
  [3],
  [20,9],
  [15,7]
]

解法：

这个题让按照二叉树的层次双数层从右至左、单数层从左至右输出；

具体实现思路如下：

  • 设置一个双端队列用于处理二叉树的结点；

  • 设置一个标记 judge 判断当前的层次的单双数

  • 对单数层进行进行 pop_front 输出，push_back 加入
  • 对双数层进行进行 pop_back 输出，push_front 加入

代码：

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/**
 * 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>> temp ;
        vector<int> cur ;
        TreeNode* now = root ;
        if(root == NULL)
        {
            return temp ;
        }
        deque<TreeNode*> q;
        q.push_back(root);
        int judge = 0 ;
        int number = 1 ;
        TreeNode* test = NULL ;
        while(q.size() != 0 )
        {
            for(int i = 0 ; i < number ; i ++)
            {
                if(judge % 2 == 0)
                {
                    test = q.back();
                    q.pop_back();
                    if(test->left)
                    {
                        q.push_front(test->left);
                    }
                    if(test->right)
                    {
                        q.push_front(test->right);
                    }
                }
                else
                {   
                    test = q.front();
                    q.pop_front();
                    if(test->right)
                    {
                        q.push_back(test->right);
                    }
                    if(test->left)
                    {
                        q.push_back(test->left);
                    }
                }   
                cur.push_back(test->val);
            }
            judge ++;
            number = q.size() ;
            temp.push_back(cur) ;
            cur.resize(0);
        }
        return temp ;
    }
};
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