题目链接:Binary Tree Zigzag Level Order Traversal
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]
]
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.
OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.
Here’s an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as “{1,2,3,#,#,4,#,#,5}”.
这道题的要求是Zigzag方式遍历二叉树,即分层遍历,先是从左到右,接着从右到左,然后从左到右,接着从右到左,以此类推。
和Binary Tree Level Order Traversal类似,由于需要把每层的节点分别放入到数组中,因此需要引入变量n记录每层的节点数量。然而由于需要按Zigzag方式遍历,因此在偶数层需要反转即可(假设根节点为第12层)。再剩下的,就是广度优先遍历的方法了。
时间复杂度:O(n)
空间复杂度:O(n)
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class Solution
{
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root)
{
vector<vector<int> > vvi;
if(root == NULL)
return vvi;
queue<TreeNode *> q;
q.push(root);
bool zigzag = false;
while(!q.empty())
{
vector<int> vi;
for(int i = 0, n = q.size(); i < n; ++ i)
{
TreeNode *temp = q.front();
q.pop();
if(temp -> left != NULL)
q.push(temp -> left);
if(temp -> right != NULL)
q.push(temp -> right);
vi.push_back(temp -> val);
}
if(zigzag)
reverse(vi.begin(), vi.end());
vvi.push_back(vi);
zigzag = !zigzag;
}
return vvi;
}
};