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.
vector<vector<int> > zigzagLevelOrder(TreeNode *root) { //C++
vector<vector<int> > result;
if(root == NULL)
return result;
int toleft = 1;
stack<TreeNode*> left;
stack<TreeNode*> right;
left.push(root);
while(!(left.empty()&&right.empty()))
{
vector<int> vec;
if(toleft==1){
while(!left.empty()){
TreeNode* temp = left.top();
vec.push_back(temp->val);
if(temp->left != NULL)
right.push(temp->left);
if(temp->right != NULL)
right.push(temp->right);
left.pop();
}
result.push_back(vec);
}
else{
while(!right.empty()){
TreeNode* temp = right.top();
vec.push_back(temp->val);
if(temp->right != NULL)
left.push(temp->right);
if(temp->left != NULL)
left.push(temp->left);
right.pop();
}
result.push_back(vec);
}
toleft *=-1;
}
return result;
}[leetcode]103 Binary Tree Zigzag Level Order Traversal
原文:http://blog.csdn.net/chenlei0630/article/details/42681253