Given a binary tree, determine if it is a complete binary tree.
Definition of a complete binary tree from Wikipedia:
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 2h nodes 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.
这道题给了一棵二叉树，让我们判断是否是一棵完全二叉树 Complete Binary Tree，通过题目中的解释可知，完全二叉树除了最后一行之外，所有位置都是满员的，而且最后一行的结点都是尽可能靠左的，注意跟完满二叉树 Full Bianry Tree 区分开来。最简单直接的方法就是按层序遍历二叉树，当遇到空结点时，后面若还出现非空结点，则一定不是完全二叉树。具体到写法就是先把根结点放入到队列中，然后进行循环，条件是队首结点不为空。在循环中取出队首结点，然后将其左右子结点加入队列中，这里不必判断子结点是否为空，为空照样加入队列，因为一旦取出空结点，循环就会停止。然后再用个循环将队首所有的空结点都移除，这样若是完全二叉树的话，队列中所有还剩的结点都应该是空结点，且都会被移除，若队列中存在非空结点，说明不是完全二叉树，最后只要判断队列是否为空即可，参见代码如下：
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();
}
};
下面这种解法思想都一样，只不过写法略有不同，这里使用了一个变量 found，初始化为 false，然后还是用层序遍历，当取出的结点为空结点时，比较 found 为 true，然后继续遍历。当遍历到非空结点时，若此时 found 为 true 了，则直接返回 false 即可。当循环退出后，返回 true， 参见代码如下：
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
参考资料：
https://leetcode.com/problems/check-completeness-of-a-binary-tree/