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52 | 52 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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53 | 53 |
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54 | 54 | ```python
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55 |
| - |
| 55 | +# Definition for a binary tree node. |
| 56 | +# class TreeNode: |
| 57 | +# def __init__(self, val=0, left=None, right=None): |
| 58 | +# self.val = val |
| 59 | +# self.left = left |
| 60 | +# self.right = right |
| 61 | +class Solution: |
| 62 | + def isCompleteTree(self, root: TreeNode) -> bool: |
| 63 | + q = deque([root]) |
| 64 | + while q: |
| 65 | + node = q.popleft() |
| 66 | + if node is None: |
| 67 | + break |
| 68 | + q.append(node.left) |
| 69 | + q.append(node.right) |
| 70 | + return all(node is None for node in q) |
56 | 71 | ```
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57 | 72 |
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58 | 73 | ### **Java**
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59 | 74 |
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60 | 75 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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61 | 76 |
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62 | 77 | ```java
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| 78 | +/** |
| 79 | + * Definition for a binary tree node. |
| 80 | + * public class TreeNode { |
| 81 | + * int val; |
| 82 | + * TreeNode left; |
| 83 | + * TreeNode right; |
| 84 | + * TreeNode() {} |
| 85 | + * TreeNode(int val) { this.val = val; } |
| 86 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 87 | + * this.val = val; |
| 88 | + * this.left = left; |
| 89 | + * this.right = right; |
| 90 | + * } |
| 91 | + * } |
| 92 | + */ |
| 93 | +class Solution { |
| 94 | + public boolean isCompleteTree(TreeNode root) { |
| 95 | + Deque<TreeNode> q = new LinkedList<>(); |
| 96 | + q.offer(root); |
| 97 | + while (q.peek() != null) { |
| 98 | + TreeNode node = q.poll(); |
| 99 | + q.offer(node.left); |
| 100 | + q.offer(node.right); |
| 101 | + } |
| 102 | + while (!q.isEmpty() && q.peek() == null) { |
| 103 | + q.poll(); |
| 104 | + } |
| 105 | + return q.isEmpty(); |
| 106 | + } |
| 107 | +} |
| 108 | +``` |
| 109 | + |
| 110 | +### **C++** |
| 111 | + |
| 112 | +```cpp |
| 113 | +/** |
| 114 | + * Definition for a binary tree node. |
| 115 | + * struct TreeNode { |
| 116 | + * int val; |
| 117 | + * TreeNode *left; |
| 118 | + * TreeNode *right; |
| 119 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 120 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 121 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 122 | + * }; |
| 123 | + */ |
| 124 | +class Solution { |
| 125 | +public: |
| 126 | + bool isCompleteTree(TreeNode* root) { |
| 127 | + queue<TreeNode*> q{{root}}; |
| 128 | + while (q.front()) |
| 129 | + { |
| 130 | + root = q.front(); |
| 131 | + q.pop(); |
| 132 | + q.push(root->left); |
| 133 | + q.push(root->right); |
| 134 | + } |
| 135 | + while (!q.empty() && !q.front()) q.pop(); |
| 136 | + return q.empty(); |
| 137 | + } |
| 138 | +}; |
| 139 | +``` |
63 | 140 |
|
| 141 | +### **Go** |
| 142 | +
|
| 143 | +```go |
| 144 | +/** |
| 145 | + * Definition for a binary tree node. |
| 146 | + * type TreeNode struct { |
| 147 | + * Val int |
| 148 | + * Left *TreeNode |
| 149 | + * Right *TreeNode |
| 150 | + * } |
| 151 | + */ |
| 152 | +func isCompleteTree(root *TreeNode) bool { |
| 153 | + q := []*TreeNode{root} |
| 154 | + for q[0] != nil { |
| 155 | + root = q[0] |
| 156 | + q = q[1:] |
| 157 | + q = append(q, root.Left) |
| 158 | + q = append(q, root.Right) |
| 159 | + } |
| 160 | + for len(q) > 0 && q[0] == nil { |
| 161 | + q = q[1:] |
| 162 | + } |
| 163 | + return len(q) == 0 |
| 164 | +} |
64 | 165 | ```
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65 | 166 |
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66 | 167 | ### **...**
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