270. Closest Binary Search Tree Value

本题我之前给想复杂了，其实非常的简单，就是道很简单的Easy题，

1. 4比3.71大，所以不用看右边，目标只可能在root和左边，测一下Root对应的值
2. 2比3.71小，所以不用看左边，目标只可能在当前的root和右边，测一下当前root对应的值
3. 3比7小

所以：

• 时间复杂度：
• 空间复杂度:

代码如下:

/**
 * 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 int closestValue(TreeNode root, double target) {
    if (root == null) {
      throw new IllegalArgumentException("Invalid Root");
    }
    int result = root.val;
    while (root != null) {
      double rootVal = (double)root.val;
      double minDiff = Math.abs((double)(result) - target);
      double diff = Math.abs(rootVal - target);
      if (diff < minDiff) {
        result = root.val;
      }
      if (rootVal < target) {
        // discard the left subtree
        root = root.right;
      }
      else {
        // discard the right subtree
        root = root.left;
      }
    }
    return result;
  }
}