[LeetCode] Binary Tree Level Order Traversal

Binary Tree Level Order Traversal

 

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 

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
]

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

解题思路：

用一个队列记录所有的节点。但是这里有个陷阱就是每一层应该分别用一个vector表示，因此还需要一个队列来存储对应的node所在的层数。因为同一层在队列中肯定是相邻的，因此可以通过这种方法来解决。

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> result;
if(root==NULL){
return result;
}
queue<int> level;
queue<TreeNode*> queues;
level.push(0);
queues.push(root);
int lastLevel = 0;
vector<int> lastValue;
while(!queues.empty()){
TreeNode* node = queues.front();
int l = level.front();
if(l!=lastLevel){
result.push_back(lastValue);
lastValue.clear();
lastLevel = l;
}
lastValue.push_back(node->val);
queues.pop();
level.pop();
if(node->left!=NULL){
queues.push(node->left);
level.push(l + 1);
}
if(node->right!=NULL){
queues.push(node->right);
level.push(l + 1);
}
}
if(!lastValue.empty()){
result.push_back(lastValue);
}
return result;
}
};