[LeetCode] Binary Tree Level Order Traversal
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Given a binary tree, return the level order traversal ofits nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7]][Solution]
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int> > res;
if(NULL == root)return res;
// insert root node
vector<vector<TreeNode*> > nodeHeap;
vector<int> rootLevVal;
vector<TreeNode*> rootLevNode;
rootLevVal.push_back(root->val);
rootLevNode.push_back(root);
res.push_back(rootLevVal);
nodeHeap.push_back(rootLevNode);
// level order
int it = 0;
int size = 1;
while(it < size){
vector<TreeNode*> tmp;
tmp = nodeHeap[it];
// level result
vector<int> levelVal;
vector<TreeNode*> levelNode;
vector<TreeNode*>::iterator i;
for(i = tmp.begin(); i != tmp.end(); ++i){
if((*i)->left != NULL){
levelVal.push_back((*i)->left->val);
levelNode.push_back((*i)->left);
}
if((*i)->right != NULL){
levelVal.push_back((*i)->right->val);
levelNode.push_back((*i)->right);
}
}
if(levelVal.size() > 0){
res.push_back(levelVal);
nodeHeap.push_back(levelNode);
size++;
}
it++;
}
return res;
}
};
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