【leetcode】Binary Tree Zigzag Level Order Traversal （middle）

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]
]

思路：由于需要排成之字形，所以用一个栈来存储当前行的内容，另一个栈来存储下一行的内容。

以根为第0层，那么偶数层都应该从左向右输出。那么每次遇到偶数层，压入下一层奇数层时就按从左到右的顺序，这样弹栈时就是从右到左。

纠结了一会儿，AC了。

class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int>> ans;
        if(root == NULL)
        {
            return ans;
        }
        int level = ; //记录当前层号
        vector<TreeNode *> curlevel;
        curlevel.push_back(root);
        while(!curlevel.empty())
        {
            vector<int> partans;
            vector<TreeNode *> nextlevel;
            if(level %  == ) //偶数层，根是第0层
            {
                while(!curlevel.empty()) //本层不为空
                {
                    if(curlevel.back()->left != NULL) //先压入左边，再压入右边
                    {
                        nextlevel.push_back(curlevel.back()->left);
                    }
                    if(curlevel.back()->right != NULL)
                    {
                        nextlevel.push_back(curlevel.back()->right);
                    }
                    partans.push_back(curlevel.back()->val);
                    curlevel.pop_back();
                }
                ans.push_back(partans);
                curlevel = nextlevel;
            }
            else
            {
                while(!curlevel.empty()) //本层不为空
                {
                    if(curlevel.back()->right != NULL) //先压入右边，再压入左边
                    {
                        nextlevel.push_back(curlevel.back()->right);
                    }
                    if(curlevel.back()->left != NULL)
                    {
                        nextlevel.push_back(curlevel.back()->left);
                    }
                    partans.push_back(curlevel.back()->val);
                    curlevel.pop_back();
                }
                ans.push_back(partans);
                curlevel = nextlevel;
            }
            level++;
        }
        return ans;
    }

    void createTree(TreeNode * &root)
    {
        int n;
        cin >> n;
        if(n != )
        {
            root = new TreeNode(n);
            createTree(root->left);
            createTree(root->right);
        }
    }
};

看看别人的答案。思路不一样。

class Solution {
vector<vector<int> > result;
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {

    if(root!=NULL)
    {
        traverse(root, );
    }

    for(int i=;i<result.size();i+=)
    {
        vector<int>* v = &result[i];
        std:reverse(v->begin(), v->end());
    }
    return result;
}

void traverse(TreeNode* node, int level)
{
    if(node == NULL) return;

    vector<int>* row = getRow(level);
    row->push_back(node->val);

    traverse(node->left, level+);
    traverse(node->right, level+);
}

vector<int>* getRow(int level)
{
    if(result.size()<=level)
    {
        vector<int> newRow;
        result.push_back(newRow);
    }
    return &result[level];
}
};