107. Binary Tree Level Order Traversal II - easy
문제
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).
제한사항
입출력 예
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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]
]
풀이
- Hash, BFS, Tree
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
if (!root) {
return {};
}
std::vector<std::vector<int>> res;
std::map<int, std::vector<int>> m;
std::queue<std::pair<TreeNode*, int>> q;
// BFS
q.push({root, 0});
while (!q.empty()) {
auto node = q.front();
q.pop();
// Map에 해당 level의 vector가 없으면 새로 추가
m[node.second].push_back(node.first->val);
if (node.first->left) {
q.push({node.first->left, node.second + 1});
}
if (node.first->right) {
q.push({node.first->right, node.second + 1});
}
}
// Map에 저장된 key(트리의 level)의 반대순으로 저장
for (auto iter = m.rbegin() ; iter != m.rend() ; ++iter) {
res.emplace_back(iter->second);
}
return res;
}
};
}