Binary Tree Level Order Traversal II
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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7]
[9,20],
[3],
]
分析
在上一题 Binary Tree Level Order Traversal 的基础上，reverse()一下即可。
递归版
// Binary Tree Level Order Traversal II
// 递归版，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int>> result;
traverse(root, 1, result);
std::reverse(result.begin(), result.end()); // 比上一题多此一行
return result;
}
void traverse(TreeNode *root, size_t level, vector<vector<int>> &result) {
if (!root) return;
if (level > result.size())
result.push_back(vector<int>());
result[level-1].push_back(root->val);
traverse(root->left, level+1, result);
traverse(root->right, level+1, result);
}
};
迭代版
// Binary Tree Level Order Traversal II
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int> > result;
if(root == nullptr) return result;
queue<TreeNode*> current, next;
vector<int> level; // elments in level level
current.push(root);
while (!current.empty()) {
while (!current.empty()) {
TreeNode* node = current.front();
current.pop();
level.push_back(node->val);
if (node->left != nullptr) next.push(node->left);
if (node->right != nullptr) next.push(node->right);
}
result.push_back(level);
level.clear();
swap(next, current);
}
reverse(result.begin(), result.end()); // 比上一题多此一行
return result;
}
};