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Description
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
- 这道题目用了BFS，进行层序遍历，用队列来实现这一过程。然后用vector装结果，由于题目要求自底向上，我们是自顶向下求的，因此我们需要把顺序翻转一下。
代码
/**
* 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>> res;
if(!root){
return res;
}
queue<TreeNode*> q;
TreeNode* tmp;
q.push(root);
while(!q.empty()){
int t=q.size();
vector<int> v;
for(int i=0;i<t;i++){
tmp=q.front();
q.pop();
v.push_back(tmp->val);
if(tmp->left){
q.push(tmp->left);
}
if(tmp->right){
q.push(tmp->right);
}
}
res.push_back(v);
}
reverse(res.begin(),res.end());
return res;
}
};