剑指 Offer 54. 二叉搜索树的第k大节点

解法 中序遍历

通过右节点->根节点->左节点的顺序即可将二叉搜索树从大到小输出。可以通过一个计数器k进行计数当k为1时就到达了目标值。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    private int k;

    public int kthLargest(TreeNode root, int k) {
        if(root == null) return 0;
        this.k = k;
        return kthLargestCore(root);
    }

    private Integer kthLargestCore(TreeNode root) {
        Integer target = null;
        if(root.right != null)
            target = kthLargestCore(root.right);
        if (target == null) {
            if (k == 1){
                target = root.val;
            }
            k--;//k为值传递不能作为函数形参
        }
        if(target == null && root.left != null) 
           target = kthLargestCore(root.left);
        return target;
    }

}