👨‍🏫 题目地址

👨‍🏫 参考题解

😋 分治

/**
 * 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 kthSmallest(TreeNode root, int k)
	{
		int n = nodeCount(root.left);//计算当前树的左子树的结点数
		if (n + 1 == k)// 说明根节点就是第 k 小的数
			return root.val;
		else if (n + 1 < k)// 说明第 k 小的数在当前树的 右子树的第（k-n-1)小处
			return kthSmallest(root.right, k - n - 1);
		else// 说明第 k 小的数在当前树的 左子树的 第（k)小处
			return kthSmallest(root.left, k);

	}

	/**
	 * @param root 当前树的根节点
	 * @return 当前树的总结点数（包含根节点）
	 */
	int nodeCount(TreeNode root)
	{
		if (root == null)
			return 0;
		return 1 + nodeCount(root.left) + nodeCount(root.right);
	}
}

😋 递归中序遍历

/**
 * 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 {
    int num = 0;
    int res;

    public int kthSmallest(TreeNode root, int k) {
        inorderTraversal(root, k);
        return res;
    }

    private void inorderTraversal(TreeNode node, int k) {
        if (node == null || num > k) {
            return;
        }
        inorderTraversal(node.left, k);
        num++;
        if (num == k) {
            res = node.val;
            return;
        }
        inorderTraversal(node.right, k);
    }
}