【LeetCode】102. Binary Tree Level Order Traversal 解題報告

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102. Binary Tree Level Order Traversal / Medium

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

Example 1:

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

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

Solution:

思路

這種橫向尋訪的題目，利用 BFS 來做會比較容易。
每次都要紀錄最右邊的節點，
而最右邊的節點則是，當目前的節點已經是之前紀錄最右邊的節點時，queue 最後面的就是下個最右邊的節點。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if(!root) return vector<vector<int>>();
        vector<vector<int>> sol(1, vector<int>());
        TreeNode* rightMost = root;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            TreeNode* cur = q.front(); q.pop();
            sol.back().emplace_back(cur->val);
            if(cur->left) q.push(cur->left);
            if(cur->right) q.push(cur->right);
            if(cur == rightMost && !q.empty()) {
                sol.emplace_back(vector<int>());
                rightMost = q.back();
            }
        }
        return sol;
    }
};