[LeetCode] 958. Check Completeness of a Binary Tree 檢查二叉樹的完全性
Given a binary tree, determine if it is acomplete binary tree.
Definition of a complete binary tree fromWikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2hnodes inclusive at the last level h.
Example 1:
Input: [1,2,3,4,5,6]
Output: true
Explanation: Every level before the last is full (ie. levels with node-values {1} and {2, 3}), and all nodes in the last level ({4, 5, 6}) are as far left as possible.
Example 2:
Input: [1,2,3,4,5,null,7] Output: false Explanation: The node with value 7 isn't as far left as possible.
Note:
- The tree will have between 1 and 100 nodes.
這道題給了一棵二叉樹，讓我們判斷是否是一棵完全二叉樹 Complete Binary Tree，通過題目中的解釋可知，完全二叉樹除了最後一行之外，所有位置都是滿員的，而且最後一行的結點都是儘可能靠左的，注意跟完滿二叉樹 Full Bianry Tree 區分開來。最簡單直接的方法就是按層序遍歷二叉樹，當遇到空結點時，後面若還出現非空結點，則一定不是完全二叉樹。具體到寫法就是先把根結點放入到佇列中，然後進行迴圈，條件是隊首結點不為空。在迴圈中取出隊首結點，然後將其左右子結點加入佇列中，這裡不必判斷子結點是否為空，為空照樣加入佇列，因為一旦取出空結點，迴圈就會停止。然後再用個迴圈將隊首所有的空結點都移除，這樣若是完全二叉樹的話，佇列中所有還剩的結點都應該是空結點，且都會被移除，若佇列中存在非空結點，說明不是完全二叉樹，最後只要判斷佇列是否為空即可，參見程式碼如下：
解法一：
class Solution {
public:
bool isCompleteTree(TreeNode* root) {
queue<TreeNode*> q{{root}};
while (q.front() != NULL) {
TreeNode *cur = q.front(); q.pop();
q.push(cur->left);
q.push(cur->right);
}
while (!q.empty() && q.front() == NULL) {
q.pop();
}
return q.empty();
}
};
下面這種解法思想都一樣，只不過寫法略有不同，這裡使用了一個變數 found，初始化為 false，然後還是用層序遍歷，當取出的結點為空結點時，比較 found 為 true，然後繼續遍歷。當遍歷到非空結點時，若此時 found 為 true 了，則直接返回 false 即可。當迴圈退出後，返回 true， 參見程式碼如下：
解法二：
class Solution {
public:
bool isCompleteTree(TreeNode* root) {
queue<TreeNode*> q{{root}};
bool found = false;
while (!q.empty()) {
TreeNode *cur = q.front(); q.pop();
if (!cur) {
found = true;
} else {
if (found) return false;
q.push(cur->left);
q.push(cur->right);
}
}
return true;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/958
參考資料：
https://leetcode.com/problems/check-completeness-of-a-binary-tree/