700. Search in a Binary Search Tree

1. Question

You are given the root of a binary search tree (BST) and an integer val.

Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

2. Examples

Example 1:

Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]

Example 2:

Input: root = [4,2,7,1,3], val = 5
Output: []

3. Constraints

• The number of nodes in the tree is in the range [1, 5000].
• 1 <= Node.val <= 10⁷
• root is a binary search tree.
• 1 <= val <= 10⁷

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/search-in-a-binary-search-tree 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

特性：左子树比根小，右子树比根大。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
  public TreeNode searchBST(TreeNode root, int val) {
    if (root == null) {
      return null;
    }
    if (root.val == val) {
      return root;
    }
    return root.val < val ? searchBST(root.right, val) : searchBST(root.left, val);

  }
}

优化：减少方法的调用，减小栈空间的使用。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
  public TreeNode searchBST(TreeNode root, int val) {
    if (root == null) {
      return null;
    }
    while (root != null) {
      if (root.val == val) {
        return root;
      } else {
        root = root.val > val ? root.left : root.right;
      }
    }
    return null;
  }
}