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,#,#,15,7},

3

/ \

9 20

/ \

15 7

return its bottom-up level order traversal as:

[

[15,7],

[9,20],

[3]
]

思路：题目要求输出二叉树层序遍历的逆序的结果。可以参考前面一题的方式，在最后的结果中将vector<vector<int>>res的数组进行reverse即可。

/**
* 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) {
queue<TreeNode*> tmp;
vector<vector<int> >res;
if(root==NULL)
return res;
tmp.push(root);
while(!tmp.empty())
{
vector<int> level_res; //save the level nu
int size=tmp.size();// 获取这个时候队列中数的个数
for(int i=0;i<size;i++)
{
TreeNode *tnode=tmp.front();
level_res.push_back(tnode->val);
tmp.pop();
if(tnode->left)
tmp.push(tnode->left);
if(tnode->right)
tmp.push(tnode->right);
}
res.push_back(level_res);
}
reverse(res.begin(),res.end());
return res;
}
};