利用二叉树层序遍历输出每层数据

107. Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:

Given binary tree 

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

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

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

本题的思路也是层序遍历，同时通过控制队列大小来控制位于同一层，可以认为与求二叉树每层平均值思路一致。
代码如下：
/**
* 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>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> v1;
vector<vector<int>> v;
vector<int> v2;
TreeNode *temp = root;
queue<TreeNode *> q;
if (temp == NULL)
return v;
q.push(temp);
v2.push_back(temp->val);
v1.push_back(v2);
v2.clear();
while (!q.empty())
{
int i = q.size();
for (int j = 0; j < i; j++)
{
TreeNode *top = q.front();
q.pop();
if (top->left != NULL)
{
q.push(top->left);
v2.push_back(top->left->
4000
val);
}

if (top->right != NULL)
{
q.push(top->right);
v2.push_back(top->right->val);
}
}
if (!v2.empty())
v1.push_back(v2);
v2.clear();
}
while (!v1.empty())
{
v.push_back(v1.back());
v1.pop_back();
}
return v;
}
};