107. Binary Tree Level Order Traversal II

Description

Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Solutions

Solution 1: BFS

We can use the BFS (Breadth-First Search) method to solve this problem. First, enqueue the root node, then continuously perform the following operations until the queue is empty:

Traverse all nodes in the current queue, store their values in a temporary array $t$, and then enqueue their child nodes.
Store the temporary array $t$ in the answer array.

Finally, return the reversed answer array.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptRustJavaScript

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans[::-1]

/**
 * 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>> ans = new LinkedList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new LinkedList<>();
        q.offerLast(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int i = q.size(); i > 0; --i) {
                TreeNode node = q.pollFirst();
                t.add(node.val);
                if (node.left != null) {
                    q.offerLast(node.left);
                }
                if (node.right != null) {
                    q.offerLast(node.right);
                }
            }
            ans.addFirst(t);
        }
        return ans;
    }
}

/**
 * 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) {
        vector<vector<int>> ans;
        if (!root) {
            return ans;
        }
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            vector<int> t;
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                t.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            ans.push_back(t);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};