Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree
Given binary tree
{3,9,20,#,#,15,7}
,3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
confused what
Analysis:
"{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
vector<vector<int> > ret;
if (root == NULL)
return ret;
queue<TreeNode*> q;
int currLevel = 1;
int nextLevel = 0;
q.push(root);
vector<int> row;
while(!q.empty())
{
TreeNode* t = q.front();
q.pop();
currLevel--;
row.push_back(t->val);
if (t->left != NULL)
{
q.push(t->left);
nextLevel++;
}
if (t->right != NULL)
{
q.push(t->right);
nextLevel++;
}
if (currLevel == 0)
{
currLevel = nextLevel;
nextLevel = 0;
ret.push_back(row);
row.clear();
}
}
return ret;
}
};
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