LeetCode 0700 Search in a Binary Search Tree【二叉搜索树】

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node’s value equals the given value. Return the subtree rooted with that node. If such node doesn’t exist, you should return NULL.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2

You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

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题意

BST 中查找等于指定值的节点

思路1

节点为空或者找到了相等的值则退出，否则在左子树或者右子树中继续查找

代码1

/**
 * 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:
    TreeNode* searchBST(TreeNode* root, int val) {
        if(root == NULL || root->val == val)
            return root;
        else if(root->val < val)
            return searchBST(root->right, val);
        else
            return searchBST(root->left, val);
    }
};