题目描述
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] ]
分析
树的层序遍历,当然是得用BFS算法。为了将每一层单独放置到一个vector容器中,可在BFS的循环体中逐层入队并逐层输出。
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> ans;
if(root==NULL) return ans;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
int cnt=q.size(); //当前层的结点数
vector<int> temp(cnt);
for(int i=0;i<cnt;i++){
auto now=q.front();
q.pop();
temp[i]=now->val;
if(now->left) q.push(now->left);
if(now->right) q.push(now->right);
}
ans.push_back(temp);
}
reverse(ans.begin(),ans.end()); //题目要求自底向上
return ans;
}
};