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32 | 32 |
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33 | 33 | <p>在上述示例中,如果要找的值是 <code>5</code>,但因为没有节点值为 <code>5</code>,我们应该返回 <code>NULL</code>。</p>
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34 | 34 |
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35 |
| - |
36 | 35 | ## 解法
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37 | 36 |
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38 | 37 | <!-- 这里可写通用的实现逻辑 -->
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44 | 43 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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45 | 44 |
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46 | 45 | ```python
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47 |
| - |
| 46 | +# Definition for a binary tree node. |
| 47 | +# class TreeNode: |
| 48 | +# def __init__(self, val=0, left=None, right=None): |
| 49 | +# self.val = val |
| 50 | +# self.left = left |
| 51 | +# self.right = right |
| 52 | +class Solution: |
| 53 | + def searchBST(self, root: TreeNode, val: int) -> TreeNode: |
| 54 | + if root is None: |
| 55 | + return None |
| 56 | + if root.val == val: |
| 57 | + return root |
| 58 | + if root.val < val: |
| 59 | + return self.searchBST(root.right, val) |
| 60 | + return self.searchBST(root.left, val) |
48 | 61 | ```
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49 | 62 |
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50 | 63 | ### **Java**
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51 | 64 |
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52 | 65 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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53 | 66 |
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54 | 67 | ```java
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| 68 | +/** |
| 69 | + * Definition for a binary tree node. |
| 70 | + * public class TreeNode { |
| 71 | + * int val; |
| 72 | + * TreeNode left; |
| 73 | + * TreeNode right; |
| 74 | + * TreeNode() {} |
| 75 | + * TreeNode(int val) { this.val = val; } |
| 76 | + * TreeNode(int val, TreeNode left, TreeNode right) { |
| 77 | + * this.val = val; |
| 78 | + * this.left = left; |
| 79 | + * this.right = right; |
| 80 | + * } |
| 81 | + * } |
| 82 | + */ |
| 83 | +class Solution { |
| 84 | + public TreeNode searchBST(TreeNode root, int val) { |
| 85 | + if (root == null) { |
| 86 | + return null; |
| 87 | + } |
| 88 | + if (root.val == val) { |
| 89 | + return root; |
| 90 | + } |
| 91 | + if (root.val < val) { |
| 92 | + return searchBST(root.right, val); |
| 93 | + } |
| 94 | + return searchBST(root.left, val); |
| 95 | + } |
| 96 | +} |
| 97 | +``` |
| 98 | + |
| 99 | +### **C++** |
| 100 | + |
| 101 | +```cpp |
| 102 | +/** |
| 103 | + * Definition for a binary tree node. |
| 104 | + * struct TreeNode { |
| 105 | + * int val; |
| 106 | + * TreeNode *left; |
| 107 | + * TreeNode *right; |
| 108 | + * TreeNode() : val(0), left(nullptr), right(nullptr) {} |
| 109 | + * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} |
| 110 | + * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} |
| 111 | + * }; |
| 112 | + */ |
| 113 | +class Solution { |
| 114 | +public: |
| 115 | + TreeNode* searchBST(TreeNode* root, int val) { |
| 116 | + if (root == nullptr) return nullptr; |
| 117 | + if (root->val == val) return root; |
| 118 | + if (root->val < val) return searchBST(root->right, val); |
| 119 | + return searchBST(root->left, val); |
| 120 | + } |
| 121 | +}; |
| 122 | +``` |
55 | 123 |
|
| 124 | +### **Go** |
| 125 | +
|
| 126 | +```go |
| 127 | +/** |
| 128 | + * Definition for a binary tree node. |
| 129 | + * type TreeNode struct { |
| 130 | + * Val int |
| 131 | + * Left *TreeNode |
| 132 | + * Right *TreeNode |
| 133 | + * } |
| 134 | + */ |
| 135 | +func searchBST(root *TreeNode, val int) *TreeNode { |
| 136 | + if root == nil { |
| 137 | + return nil |
| 138 | + } |
| 139 | + if root.Val == val { |
| 140 | + return root |
| 141 | + } |
| 142 | + if root.Val < val { |
| 143 | + return searchBST(root.Right, val) |
| 144 | + } |
| 145 | + return searchBST(root.Left, val) |
| 146 | +} |
56 | 147 | ```
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57 | 148 |
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58 | 149 | ### **...**
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