[LeetCode] 919. Complete Binary Tree Inserter 徹底二叉樹插入器

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.

這道題說是讓實現一個徹底二叉樹的插入器的類，以前也作過關於徹底二叉樹的題 Count Complete Tree Nodes。首先須要搞清楚的是徹底二叉樹的定義，即對於一顆二叉樹，假設其深度爲d（d>1）。除了第d層外，其它各層的節點數目均已達最大值，且第d層全部節點從左向右連續地緊密排列，換句話說，徹底二叉樹從根結點到倒數第二層知足完美二叉樹，最後一層能夠不徹底填充，其葉子結點都靠左對齊。因爲插入操做要找到最後一層的第一個空缺的位置，因此很天然的就想到了使用層序遍歷的方法，因爲插入函數返回的是插入位置的父結點，因此在層序遍歷的時候，只要遇到某個結點的左子結點或者右子結點不存在，則跳出循環，則這個殘缺的父結點恰好就在隊列的首位置。那麼在插入函數時，只要取出這個殘缺的父結點，判斷若其左子結點不存在，說明新的結點要鏈接在左子結點上，不然將新的結點鏈接在右子結點上，並把此時的左右子結點都存入隊列中，並將以前的隊首元素移除隊列便可，參見代碼以下：

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;
};

下面這種解法縮短了建樹的時間，可是極大的增長了插入函數的運行時間，由於每插入一個結點，都要從頭開始再遍歷一次，並非很高效，能夠看成一種發散思惟吧，參見代碼以下：

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;
};

再來看一種不使用隊列的解法，由於隊列老是要遍歷，比較麻煩，若是使用數組來按層序遍歷的順序保存這個徹底二叉樹的結點，將會變得十分的簡單。並且有個最大的好處是，能夠直接經過座標定位到其父結點的位置，經過 (i-1)/2 來找到父結點，這樣的話就完美的解決了插入函數要求返回父結點的要求，並且經過判斷當前完整二叉樹結點個數的奇偶，能夠得知最後一個結點是在左子結點上仍是右子結點上，這樣就能夠直接將新加入的結點連到到父結點的正確的子結點位置，參見代碼以下：

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:

https://github.com/grandyang/leetcode/issues/919

相似題目：

Count Complete Tree Nodes

參考資料：

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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