题目地址：https://leetcode.com/problems/check-completeness-of-a-binary-tree/
题目描述
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:
1、 Thetreewillhavebetween1and100nodes.；
题目大意
判断一个二叉树是不是完全二叉树。
解题方法
BFS
这个题可以使用DFS或者BFS先找出二叉树的层次遍历。之后的判断中，使用DFS比较麻烦一些。
使用BFS的话层次遍历比较简单，因为我们从每层的从左到右进行遍历，如果某一层已经出现None之后，后面还有非空叶子节点的话，那么就不是完全二叉树。
代码如下：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isCompleteTree(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root: return True
res = []
que = collections.deque()
que.append(root)
hasNone = False
while que:
size = len(que)
for i in range(size):
node = que.popleft()
if not node:
hasNone = True
continue
if hasNone:
return False
que.append(node.left)
que.append(node.right)
return True
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C++代码如下：
/**
* 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:
bool isCompleteTree(TreeNode* root) {
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
TreeNode* node = q.front(); q.pop();
if (!node) break;
q.push(node->left);
q.push(node->right);
}
while (!q.empty()) {
TreeNode* node = q.front(); q.pop();
if (node)
return false;
}
return true;
}
};
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DFS
思路是，除了最后一层之外，其余的层必须都是满二叉树，最后一层左边只能全部是非空叶子节点，如果出现None之后，后面不能再有非空叶子节点了。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isCompleteTree(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if not root: return True
res = []
self.getlevel(res, 0, root)
depth = len(res) - 1
for d in range(depth):
if d != depth - 1:
if None in res[d] or len(res[d]) != (2 ** d):
return False
else:
ni = -1
for i, n in enumerate(res[d]):
if n == None:
if ni == -1:
ni = i
else:
if ni != -1:
return False
return True
def getlevel(self, res, level, root):
if level >= len(res):
res.append([])
if not root:
res[level].append(None)
else:
res[level].append(root.val)
self.getlevel(res, level + 1, root.left)
self.getlevel(res, level + 1, root.right)
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