[LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二

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]
]

从底部层序遍历其实还是从顶部开始遍历，只不过最后存储的方式有所改变，可以参见我之前的博文 /article/4901519.html。 代码如下：

解法一：

// Iterative
class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int> > res;
if (root == NULL) return res;

queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
vector<int> oneLevel;
int size = q.size();
for (int i = 0; i < size; ++i) {
TreeNode *node = q.front();
q.pop();
oneLevel.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
res.insert(res.begin(), oneLevel);
}
return res;
}
};

下面我们来看递归的解法，核心就在于我们需要一个二维数组，和一个变量level，当level递归到上一层的个数，我们新建一个空层，继续往里面加数字，参见代码如下：

解法二：

// Recurive
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int> > res;
levelorder(root, 0, res);
return vector<vector<int> > (res.rbegin(), res.rend());
}
void levelorder(TreeNode *root, int level, vector<vector<int> > &res) {
if (!root) return;
if (res.size() == level) res.push_back({});
res[level].push_back(root->val);
if (root->left) levelorder(root->left, level + 1, res);
if (root->right) levelorder(root->right, level + 1, res);
}
};

类似题目：

Binary Tree Zigzag Level Order Traversal

Binary Tree Level Order Traversal

LeetCode All in One 题目讲解汇总(持续更新中...)