题目
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] ]
层序遍历,反向输出结果(自底向上)。
bfs,然后反转结果即可。
代码:
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
struct nd //探测用点
{
TreeNode *tn; //位置
int deep; //深度
nd(TreeNode *t=NULL,int d=0):tn(t),deep(d){}
};
vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int>> ret; //返回值
deque<nd> queue(1,nd(root)); //队列
while(!queue.empty()) //bfs
{
if(queue.front().tn!=NULL)
{
if(queue.front().deep>=ret.size())
ret.push_back(vector<int>(0,0));
ret[queue.front().deep].push_back(queue.front().tn->val);
queue.push_back(nd(queue.front().tn->left,queue.front().deep+1));
queue.push_back(nd(queue.front().tn->right,queue.front().deep+1));
}
queue.pop_front();
}
reverse(ret.begin(),ret.end()); //反转
return ret;
}
};