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

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+# Leetcode Binary-Tree-Level-Order-Traversal
+
+#### 2022-07-05 09:09
+
+> ##### Algorithms:
+> #algorithm #BFS
+> ##### Data structures:
+> #DS #binary_tree 
+> ##### Difficulty:
+> #coding_problem #difficulty-medium
+> ##### Additional tags:
+> #leetcode
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#BFS
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/binary-tree-level-order-traversal/)
+___
+### 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:**
+
+![](https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg)
+
+**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.
+
+```cpp
+vector<vector<int>> vec;
+vec.push_back({});
+vec.back().push_back(5);
+// [[5]]
+```
+### Solution
+
+```cpp
+/**
+ * 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;
+  }
+};
+```