107

107. Binary Tree Level Order Traversal II (Medium)

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

Companies:
Amazon, Apple

Related Topics:
Tree, Breadth-First Search, Binary Tree

Similar Questions:

• Binary Tree Level Order Traversal (Medium)
• Average of Levels in Binary Tree (Easy)

Solution 1. BFS

// OJ: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        if (!root) return {};
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int cnt = q.size();
            ans.emplace_back();
            while (cnt--) {
                root = q.front();
                q.pop();
                ans.back().push_back(root->val);
                if (root->left) q.push(root->left);
                if (root->right) q.push(root->right);
            }
        }
        reverse(begin(ans), end(ans));
        return ans;
    }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    vector<vector<int>> ans;
    void dfs(TreeNode *root, int lv) {
        if (!root) return;
        if (lv >= ans.size()) ans.emplace_back();
        ans[lv].push_back(root->val);
        dfs(root->left, lv + 1);
        dfs(root->right, lv + 1);
    }
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        dfs(root, 0);
        reverse(begin(ans), end(ans));
        return ans;
    }
};