LeetCode-in-All
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
#include <vector>
#include <queue>
/**
* 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>> levelOrder(TreeNode* root) {
vector<vector<int>> ans;
if (!root) return ans;
queue<pair<TreeNode*, int>> q;
int level= 0;
q.push({root, level});
while (!q.empty()) {
pair<TreeNode*, int> temp = q.front();
q.pop();
if (temp.second >= ans.size()) {
ans.push_back({temp.first->val});
}
else {
ans[temp.second].push_back(temp.first->val);
}
level = temp.second + 1;
if (temp.first->left) q.push({temp.first->left, level});
if (temp.first->right) q.push({temp.first->right, level});
}
return ans;
}
};