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

Code

Time Complexity: $O(n)$, Space Complexity: $O(n)$

先做 Binary Tree Level Order Traversal，然後將結果 reverse 即可。

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

Source

• Binary Tree Level Order Traversal II - LeetCode