二叉搜索树的第k个结点

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题目描述：给定一颗二叉搜索树，请找出其中的第k大的结点。例如， 5 / \ 3 7 /\ /\ 2 4 6 8 中，按结点数值大小顺序第三个结点的值为4。

public class TreeNode {    int val = 0;    TreeNode left = null;    TreeNode right = null;    public TreeNode(int val) {        this.val = val;    }}

思路一：

中序遍历，即从小到大排序

import java.util.ArrayList;public class Solution {    ArrayList<Integer> list = new ArrayList<>();    TreeNode KthNode(TreeNode pRoot, int k)    {        if (pRoot == null || k <= 0)            return null;        inorder(pRoot);        if (k > list.size())            return null;        return new TreeNode(list.get(k - 1));    }    private void inorder(TreeNode root)    {        if (root == null)            return;        inorder(root.left);        list.add(root.val);        inorder(root.right);    }}

思路二：

中序遍历

public class Solution {    int index = 0;    TreeNode KthNode(TreeNode pRoot, int k)    {        if (pRoot != null)        {            TreeNode node = KthNode(pRoot.left, k);            if (node != null)                return node;            index++;            if (index == k)                return pRoot;            node = KthNode(pRoot.right, k);            if (node != null)                return node;        }        return null;    }}

思路三：

中序遍历，非递归

import java.util.Stack;public class Solution {    TreeNode KthNode(TreeNode pRoot, int k)    {        if (pRoot == null || k <= 0)            return null;        Stack<TreeNode> stack = new Stack<>();        int count = 0;        TreeNode node = pRoot;        while (node != null || !stack.isEmpty())        {            if (node != null)            {                stack.push(node);                node = node.left;            }            else             {                node = stack.pop();                count++;                if (count == k)                    return node;                node = node.right;            }        }        return null;    }}

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