剑指Offer-54：二叉搜索树的第k个结点

题目：二叉搜索树的第k个结点
给定一棵二叉搜索树，按节点值从小到大排列，找出其中的第k（k≥1）个结点。

主要思路：使用中序遍历，每遍历一个节点时，判断k的值，若k等于1，则当前遍历到的节点就是所求节点，否则使k减1。

关键点：中序遍历

时间复杂度：O(n)

public class KthNodeOfTree{
    private static TreeNode result;
    private static int runK;

    public static void main(String[] args){
//            10
//         /      \
//        6        14
//       /\        /\
//      4  8     12  16
        TreeNode root = TreeNode.generateBinaryTree();
        System.out.println(findKthNode(root, 1).val);  //4
        System.out.println(findKthNode(root, 3).val);  //8
        System.out.println(findKthNode(root, 5).val);  //12
        //System.out.println(findKthNodeByLoop(root, 5).val);  //12
    }

    private static TreeNode findKthNode(TreeNode root, int k) {
        if (root == null || k == 0) return null;
        result = null;
        runK = k;
        findKthNode(root);
        return result;
    }

    //递归
    private static void findKthNode(TreeNode root){
        if (root == null || result != null) return;
        //中序遍历，先遍历左节点
        findKthNode(root.left);
        if (result == null) {
            if (runK == 1){
                result = root;
                return;
            }
            runK--;
        }
        //遍历右节点
        findKthNode(root.right);
    }

    //循环
    private static TreeNode findKthNodeByLoop(TreeNode root, int k){
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            if (!stack.isEmpty()){
                cur = stack.pop();
                if (k == 1){
                    return cur;
                }
                k--;
                cur = cur.right;
            }
        }
        return null;
    }
}

我的版本：

 	int sum = 0;
    TreeNode ret = null;
    TreeNode KthNode(TreeNode pRoot, int k){
    	if(pRoot == null)
    		return ret;
        if(pRoot.left != null)
        	KthNode(pRoot.left, k);
        sum++;
        if(sum == k) {
        	ret =  pRoot;
        }
        if(pRoot.right != null)
        	KthNode(pRoot.right, k);
        return ret;
    }

难理解版递归：

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;
    }
}