LeetCode 102. Binary Tree Level Order Traversal & LeetCode 107. Binary Tree Level Order Traversal II

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博客围绕二叉树层次遍历展开，介绍了102题二叉树层次遍历及107题自底向上层次遍历。对于102题，给出了BFS和DFS两种解法，BFS用队列存储节点，DFS类似前序遍历采用递归实现。107题则是将102题输出结果反转。

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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],

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],

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;
    }
};