LeetCode OJ 之 Binary Tree Level Order Traversal （二叉树的层次遍历）

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]
]

思路：

定义两个队列，分别存储当前行的结点和下一行的结点，通过入队出队可以按行遍历结点数据，并存入vector。

代码：

迭代版：

/**
* Definition for binary tree
* 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 == nullptr)
return result;
queue<TreeNode*> current, next;//定义两个队列，current存当前行的结点，next存储下一行的结点
vector<int> level; //存储当前行的数据
current.push(root);//把根结点入队
while (!current.empty())
{
while (!current.empty()) //结束条件是当前队列为空，下一队列非空
{
TreeNode* node = current.front();
current.pop();
level.push_back(node->val);
if (node->left != nullptr) next.push(node->left);//把当前行结点的子结点都存入next队列
if (node->right != nullptr) next.push(node->right);
}
result.push_back(level);//把当前行的vector push进result
level.clear();//清空当前vector里的数据
swap(next, current);//交换两个队列后，下一行的结点存在current里，next则为空
}
return result;
}
};

递归版：

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

class Solution {
public:
void traverse(TreeNode *root, size_t level, vector<vector<int>> &result)
{
if (!root)
return;
if (level > result.size())
result.push_back(vector<int>());

result[level-1].push_back(root->val);

traverse(root->left, level+1, result);
traverse(root->right, level+1, result);

}
vector<vector<int> > levelOrder(TreeNode *root)
{
vector<vector<int>> result;
traverse(root, 1, result);
return result;
}
};