A  complete  binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible.

Write a data structure CBTInserter that is initialized with a complete binary tree and supports the following operations:

• CBTInserter(TreeNode root) initializes the data structure on a given tree with head node root;
• CBTInserter.insert(int v) will insert a TreeNode into the tree with value node.val = v so that the tree remains complete, and returns the value of the parent of the inserted TreeNode;
• CBTInserter.get_root() will return the head node of the tree.

Example 1:

Input: inputs = ["CBTInserter","insert","get_root"], inputs = [[[1]],[2],[]]
Output: [null,1,[1,2]]

Example 2:

Input: inputs = ["CBTInserter","insert","insert","get_root"], inputs = [[[1,2,3,4,5,6]],[7],[8],[]]
Output: [null,3,4,[1,2,3,4,5,6,7,8]]

Note:

1. The initial given tree is complete and contains between 1 and 1000 nodes.
2. CBTInserter.insert is called at most 10000 times per test case.
3. Every value of a given or inserted node is between 0 and 5000.

这道题说是让实现一个完全二叉树的插入器的类，之前也做过关于完全二叉树的题 Count Complete Tree Nodes。首先需要搞清楚的是完全二叉树的定义，即对于一颗二叉树，假设其深度为d（d>1）。除了第d层外，其它各层的节点数目均已达最大值，且第d层所有节点从左向右连续地紧密排列，换句话说，完全二叉树从根结点到倒数第二层满足完美二叉树，最后一层可以不完全填充，其叶子结点都靠左对齐。由于插入操作要找到最后一层的第一个空缺的位置，所以很自然的就想到了使用层序遍历的方法，由于插入函数返回的是插入位置的父结点，所以在层序遍历的时候，只要遇到某个结点的左子结点或者右子结点不存在，则跳出循环，则这个残缺的父结点刚好就在队列的首位置。那么在插入函数时，只要取出这个残缺的父结点，判断若其左子结点不存在，说明新的结点要连接在左子结点上，否则将新的结点连接在右子结点上，并把此时的左右子结点都存入队列中，并将之前的队首元素移除队列即可，参见代码如下：

解法一：

class CBTInserter {
public:
    CBTInserter(TreeNode* root) {
        tree_root = root;
        q.push(root);
        while (!q.empty()) {
            auto t = q.front(); 
            if (!t->left || !t->right) break;
            q.push(t->left);
            q.push(t->right);
            q.pop();
        }
    }   
    int insert(int v) {
        TreeNode *node = new TreeNode(v);
        auto t = q.front(); 
        if (!t->left) t->left = node;
        else {
            t->right = node;
            q.push(t->left);
            q.push(t->right);
            q.pop();
        }
        return t->val;
    }  
    TreeNode* get_root() {
        return tree_root;
    }

private:
    TreeNode *tree_root;
    queue<TreeNode*> q;
};

下面这种解法缩短了建树的时间，但是极大的增加了插入函数的运行时间，因为每插入一个结点，都要从头开始再遍历一次，并不是很高效，可以当作一种发散思维吧，参见代码如下：

解法二：

class CBTInserter {
public:
    CBTInserter(TreeNode* root) {
        tree_root = root;
    }
    int insert(int v) {
        queue<TreeNode*> q{{tree_root}};
        TreeNode *node = new TreeNode(v);
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            if (t->left) q.push(t->left);
            else {
                t->left = node;
                return t->val;
            }
            if (t->right) q.push(t->right);
            else {
                t->right = node;
                return t->val;
            }
        }
        return 0;
        
    }    
    TreeNode* get_root() {
        return tree_root;
    }

private:
    TreeNode *tree_root;
};

再来看一种不使用队列的解法，因为队列总是要遍历，比较麻烦，如果使用数组来按层序遍历的顺序保存这个完全二叉树的结点，将会变得十分的简单。而且有个最大的好处是，可以直接通过坐标定位到其父结点的位置，通过 (i-1)/2 来找到父结点，这样的话就完美的解决了插入函数要求返回父结点的要求，而且通过判断当前完整二叉树结点个数的奇偶，可以得知最后一个结点是在左子结点上还是右子结点上，这样就可以直接将新加入的结点连到到父结点的正确的子结点位置，参见代码如下：

解法三：

class CBTInserter {
public:
    CBTInserter(TreeNode* root) {
        tree.push_back(root);
        for (int i = 0; i < tree.size(); ++i) {
            if (tree[i]->left) tree.push_back(tree[i]->left);
            if (tree[i]->right) tree.push_back(tree[i]->right);
        }
    }
    int insert(int v) {
        TreeNode *node = new TreeNode(v);
        int n = tree.size();
        tree.push_back(node);
        if (n % 2 == 1) tree[(n - 1) / 2]->left = node;
        else tree[(n - 1) / 2]->right = node;
        return tree[(n - 1) / 2]->val;
    }    
    TreeNode* get_root() {
        return tree[0];
    }

private:
    vector<TreeNode*> tree;
};

Github 同步地址:

#919

类似题目：

Count Complete Tree Nodes

参考资料：

https://leetcode.com/problems/complete-binary-tree-inserter/

https://leetcode.com/problems/complete-binary-tree-inserter/discuss/178424/C%2B%2BJavaPython-O(1)-Insert

https://leetcode.com/problems/complete-binary-tree-inserter/discuss/178528/Java-Solution%3A-O(1)-Insert-VS.-O(1)-Pre-process-Trade-Off

https://leetcode.com/problems/complete-binary-tree-inserter/discuss/178427/Java-BFS-straightforward-code-two-methods-Initialization-and-insert-time-O(1)-respectively.

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