102. Binary Tree Level Order Traversal
题目：
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example: Given binary tree [3,9,20,null,null,15,7],
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]这道题的本质就是层次遍历二叉树，并按层打印出所有节点。感觉和之前的maximum depth的BFS做法是一样的，使用queue来存储节点。由于要求按层打印，于是在对每一层遍历的过程中，设立一个临时的vector对该层节点进行存储即可，还是比较简单的。代码如下，时间100%，空间43.96%：
/* * @lc app=leetcode id=102 lang=cpp * * [102] Binary Tree Level Order Traversal */ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> result; if (!root) { return result; } queue<TreeNode*> nodes; nodes.push(root); while (!nodes.empty()) { int size = nodes.size(); vector<int> temp; for (int i = 0; i < size; i++) { TreeNode* front = nodes.front(); nodes.pop(); temp.push_back(front->val); if (front->left) { nodes.push(front->left); } if (front->right) { nodes.push(front->right); } } result.push_back(temp); } return result; } };2020.09.27 Java:
Runtime: 1 ms, faster than 62.18% of Java online submissions for Binary Tree Level Order Traversal.
Memory Usage: 39.2 MB, less than 98.57% of Java online submissions for Binary Tree Level Order Traversal.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<List<Integer>> levelOrder(TreeNode root) { if (root == null) { return new ArrayList<>(); } Queue<TreeNode> queue = new ArrayDeque<>(); queue.add(root); List<List<Integer>> result = new ArrayList<>(); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> level = new ArrayList<>(); for (int i = 0; i < size; i++) { TreeNode node = queue.poll(); if (node.left != null) { queue.add(node.left); } if (node.right != null) { queue.add(node.right); } level.add(node.val); } result.add(level); } return result; } }除了常规的层次遍历BFS以外，还有一种DFS的骚操作，看discussion里说是类似于前序遍历，采用递归实现。在递归函数的参数中，还需要传递当前所在的层的层数，用于判断递归的次数和把节点的值塞到对应的层数中。在递归函数中，递归基是当根节点为空时；在其他条件下，需要通过层数和结果vector的数值来判断是否为第一次遍历，也就是刚开始的向下遍历，可以算是用于计算层数的部分吧，有多少层就要push多少个层的vector进去，并且把当前节点的值塞到对应的层数中，然后往左边遍历，最后往右边遍历，确实也就是前序遍历的顺序了。代码如下，时间93.09%，空间8.51%：
/* * @lc app=leetcode id=102 lang=cpp * * [102] Binary Tree Level Order Traversal */ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> result; void dfs(TreeNode* root, int depth) { if (!root) { return; } if (depth == result.size()) { result.push_back(vector<int>()); } result[depth].push_back(root->val); dfs(root->left, depth + 1); dfs(root->right, depth + 1); } vector<vector<int>> levelOrder(TreeNode* root) { dfs(root, 0); return result; } };2020.09.27 Java：
DFS的思想在于通过把层数作为一个参数，并通过检查result的大小来判断是否之前已经遍历过这个层的节点，如果没遍历过（即层数=result.size()）则建一个新层，如果遍历过就取出对应的list并把当前节点加进去。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Binary Tree Level Order Traversal.
Memory Usage: 39.3 MB, less than 95.93% of Java online submissions for Binary Tree Level Order Traversal.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<>(); dfs(root, 0, result); return result; } private void dfs(TreeNode root, int level, List<List<Integer>> result) { if (root == null) { return; } if (level == result.size()) { result.add(new ArrayList<>()); } List<Integer> levelList = result.get(level); levelList.add(root.val); dfs(root.left, level + 1, result); dfs(root.right, level + 1, result); } }107. Binary Tree Level Order Traversal II
题目：
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example: Given binary tree [3,9,20,null,null,15,7],
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]这道题的本质就是把前面一道题的输出给倒过来。看了一圈discussion好像没什么其他的insight，于是就重新复习了前面一道题的层次遍历两种方法，最后把vector给reverse一下。除了巩固毫无收获的一道题，就和之前的放在一起了。下面贴代码：
BFS：
/* * @lc app=leetcode id=107 lang=cpp * * [107] Binary Tree Level Order Traversal II */ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode* root) { vector<vector<int>> result; if (!root) { return result; } queue<TreeNode*> nodes; nodes.push(root); while (!nodes.empty()) { int size = nodes.size(); vector<int> temp; for (int i = 0; i < size; i++) { TreeNode* front = nodes.front(); nodes.pop(); temp.push_back(front->val); if (front->left) { nodes.push(front->left); } if (front->right) { nodes.push(front->right); } } result.push_back(temp); } reverse(result.begin(), result.end()); return result; } };DFS：
/* * @lc app=leetcode id=107 lang=cpp * * [107] Binary Tree Level Order Traversal II */ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> result; void dfs(TreeNode* root, int depth) { if (!root) { return; } if (depth == result.size()) { result.push_back(vector<int>()); } result[depth].push_back(root->val); dfs(root->left, depth + 1); dfs(root->right, depth + 1); } vector<vector<int>> levelOrderBottom(TreeNode* root) { dfs(root, 0); reverse(result.begin(), result.end()); return result; } };
