logseq_notes/pages/OJ notes/pages/Leetcode Binary-Tree-Level-Order-Traversal.md

Leetcode Binary-Tree-Level-Order-Traversal

2022-07-05 09:09

Algorithms:

#algorithm #BFS

Data structures:

#DS #binary_tree

Difficulty:

#coding_problems #difficulty_medium

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

  ---

Problem

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).

Examples

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

Thoughts

[!summary] This is a #BFS problem.

In contrary to DFS, BFS uses queue. and there are many tricks for pushing 2d arrays.

vector<vector<int>> vec;
vec.push_back({});
vec.back().push_back(5);
// [[ 5 ]]

Solution

/**
 * 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) {
  // Using queue
  queue<TreeNode *> pending;
  vector<vector<int>> answer;

  TreeNode *ptr = root;
  if (ptr)
    pending.push(ptr);

  while (!pending.empty()) {
    answer.push_back({});
    // After each while loop, every element in queue is in next level.
    for (int i = 0, size = pending.size(); i < size; i++) {
      ptr = pending.front();
      pending.pop();

      if (ptr->left)
        pending.push(ptr->left);
      if (ptr->right)
        pending.push(ptr->right);

      answer.back().push_back(ptr->val);
    }
  }

  return answer;
}
};