102. Binary Tree Level Order Traversal
题目
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.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
解析
// 102. Binary Tree Level Order Traversal
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
vector<vector<int>> vecs;
if (!root)
{
return vecs;
}
queue<TreeNode*> que;
que.push(root);
while (!que.empty())
{
int size = que.size();
vector<int> vec;
while (size--)
{
TreeNode* temp = que.front();
que.pop();
vec.push_back(temp->val);
if (temp->left)
{
que.push(temp->left);
}
if (temp->right)
{
que.push(temp->right);
}
}
vecs.push_back(vec);
}
return vecs;
}
vector<vector<int>> levelOrder(TreeNode* root) {
if (!root) { return{}; }
vector<int> row;
vector<vector<int> > result;
queue<TreeNode*> q;
q.push(root);
int count = 1;
while (!q.empty()) {
if (q.front()->left) { q.push(q.front()->left); }
if (q.front()->right) { q.push(q.front()->right); }
row.push_back(q.front()->val), q.pop();
if (--count == 0) {
result.emplace_back(row), row.clear();
count = q.size();
}
}
return result;
}
// 递归实现;up->bottom
//类似的：bottom->up: 107. Binary Tree Level Order Traversal II: http://www.cnblogs.com/ranjiewen/p/8253217.html
vector<vector<int>> ret;
void buildVector(TreeNode *root, int depth)
{
if (root == NULL) return;
if (ret.size() == depth)
ret.push_back(vector<int>());
ret[depth].push_back(root->val);
buildVector(root->left, depth + 1);
buildVector(root->right, depth + 1);
}
vector<vector<int> > levelOrder(TreeNode *root) {
buildVector(root, 0);
return ret;
}
};
题目来源
