题目：给定一棵二叉搜索树，请找出当中的第k大的结点。

解题思路

　　假设依照中序遍历的顺序遍历一棵二叉搜索树，遍历序列的数值是递增排序的。

仅仅须要用中序遍历算法遍历一棵二叉搜索树。就非常easy找出它的第k大结点。

结点定义

private static class BinaryTreeNode {

    private int val;

    private BinaryTreeNode left;

    private BinaryTreeNode right;

    public BinaryTreeNode() {

    }

    public BinaryTreeNode(int val) {

        this.val = val;

    }

    @Override

    public String toString() {

        return val + "";

    }

}

代码实现

public class Test63 {

    private static class BinaryTreeNode {

        private int val;

        private BinaryTreeNode left;

        private BinaryTreeNode right;

        public BinaryTreeNode() {

        }

        public BinaryTreeNode(int val) {

            this.val = val;

        }

        @Override

        public String toString() {

            return val + "";

        }

    }

    public static BinaryTreeNode kthNode(BinaryTreeNode root, int k) {

        if (root == null || k < 1) {

            return null;

        }

        int[] tmp = {k};

        return kthNodeCore(root, tmp);

    }

    private static BinaryTreeNode kthNodeCore(BinaryTreeNode root, int[] k) {

        BinaryTreeNode result = null;

        // 先成左子树中找

        if (root.left != null) {

          result =  kthNodeCore(root.left, k);

        }

        // 假设在左子树中没有找到

        if (result == null) {

            // 说明当前的根结点是所要找的结点

            if(k[0] == 1) {

                result = root;

            } else {

                // 当前的根结点不是要找的结点。可是已经找过了。所以计数器减一

                k[0]--;

            }

        }

        // 根结点以及根结点的左子树都没有找到，则找其右子树

        if (result == null && root.right != null) {

            result = kthNodeCore(root.right, k);

        }

        return result;

    }

    public static void main(String[] args) {

        BinaryTreeNode n1 = new BinaryTreeNode(1);

        BinaryTreeNode n2 = new BinaryTreeNode(2);

        BinaryTreeNode n3 = new BinaryTreeNode(3);

        BinaryTreeNode n4 = new BinaryTreeNode(4);

        BinaryTreeNode n5 = new BinaryTreeNode(5);

        BinaryTreeNode n6 = new BinaryTreeNode(6);

        BinaryTreeNode n7 = new BinaryTreeNode(7);

        BinaryTreeNode n8 = new BinaryTreeNode(8);

        BinaryTreeNode n9 = new BinaryTreeNode(9);

        n1.left = n2;

        n1.right = n3;

        n2.left = n4;

        n2.right = n5;

        n3.left = n6;

        n3.right = n7;

        n4.left = n8;

        n4.right = n9;

        print(n1);

        System.out.println();

        for (int i = 0; i <= 10; i++) {

            System.out.printf(kthNode(n1, i) + ", ");

        }

    }

    /**

     * 中序遍历一棵树

     * @param root

     */

    private static void print(BinaryTreeNode root) {

        if (root != null) {

            print(root.left);

            System.out.printf("%-3d", root.val);

            print(root.right);

        }

    }

}

执行结果

特别说明

欢迎转载，转载请注明出处【http://blog.csdn.net/DERRANTCM/article/details/46872313】