Leetcode 102 Binary Tree Level Order Traversal 二叉树+BFS

/**
* 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>> v;
if(!root) return v;
vector<int> t;

t.push_back(root->val);
v.push_back(t);
root->val = 1;

queue<TreeNode*> q;
q.push(root);

while(!q.empty()){
TreeNode* now = q.front();
q.pop();
if(!now) continue;

q.push(now->left);
q.push(now->right);
if(now->val < v.size()){
if(now->left) v[now->val].push_back(now->left->val);
if(now->right) v[now->val].push_back(now->right->val);
}
else{
vector<int> t;
if(now->left) t.push_back(now->left->val);
if(now->right) t.push_back(now->right->val);
if(t.size() != 0)v.push_back(t);
}
if(now->left) now->left->val = now->val + 1;
if(now->right)now->right->val = now->val + 1;
}
//reverse(v.begin(),v.end());
return v;
}
};