题目描述：
给定一个二叉树，返回其节点值自底向上的层序遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）
例如：
给定二叉树 [3,9,20,null,null,15,7],
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
解题思路：
- 每次都将二叉树结点插入到队列的首位，这样先进入队列的结点最后输出。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> arr = new LinkedList<List<Integer>>();
if (root == null) return arr;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
List<Integer> ans = new LinkedList<Integer>();
int len = queue.size();
for (int i = 0; i < len; i++) {
TreeNode cur = queue.poll();
ans.add(cur.val);
if (cur.left != null) {
queue.offer(cur.left);
}
if (cur.right != null) {
queue.offer(cur.right);
}
}
arr.addFirst(ans);
}
return arr;
}
}
arr.addFirst(ans) 相当于arr.add(0, ans)，还可以写成这样。
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> arr = new LinkedList<List<Integer>>();
if (root == null) return arr;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
List<Integer> ans = new LinkedList<Integer>();
int len = queue.size();
for (int i = 0; i < len; i++) {
TreeNode cur = queue.poll();
ans.add(cur.val);
if (cur.left != null) {
queue.offer(cur.left);
}
if (cur.right != null) {
queue.offer(cur.right);
}
}
arr.add(0, ans);
}
return arr;
}
}
如果用C++语言编写程序，直接用翻转函数reverse(result.begin(), result.end());代码还可以写成下面这样
/**
* 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) {
queue<TreeNode*> que;
if (root != NULL) que.push(root);
vector<vector<int>> result;
while (!que.empty()) {
int size = que.size();
vector<int> vec;
for (int i = 0; i < size; i++) {
TreeNode* node = que.front();
que.pop();
vec.push_back(node->val);
if (node->left) que.push(node->left);
if (node->right) que.push(node->right);
}
result.push_back(vec);
}
reverse(result.begin(), result.end());
return result;
}
};