binary-tree-zigzag-level-order-traversal——二叉树分层输出
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.
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}".
/**
* 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>> res;
if(root==NULL) return res;
queue<TreeNode*> q;
q.push(root);
bool reverse=false;
while(!q.empty()){
vector<int> v;
int size=q.size();
for(int i=;i<size;++i){
TreeNode *cur=q.front();
q.pop();
v.push_back(cur->val);
if(cur->left!=NULL) q.push(cur->left);
if(cur->right!=NULL) q.push(cur->right);
}
if(reverse){ vector<int> tmp;
for(int i=v.size()-;i>=;--i){
tmp.push_back(v[i]);
}
res.push_back(tmp);
}else{
res.push_back(v);
}
reverse=!reverse;
}
return res;
}
};
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