二叉查找树其他操作

将二叉搜索树转变为排序的双向链表

来源：剑指offer27
思路：想成是中序遍历的变形，可以利用中序遍历的非递归方式不断压栈弹栈，保存前一个节点与其互指。
也可以使用递归方式，每次可以返回已成形链表的最后一个节点，用给节点连接当前节点，再将当前节点的右子树的最小节点连接起来。

public TreeNode last = null;public TreeNode Convert(TreeNode root){    if(root == null) return null;    TreeNode head = root;    for(; head.left != null; head = head.left);    ConvertToList(root);    return head;}public void ConvertToList(TreeNode root){    if(root.left != null) ConvertToList(root.left);    root.left = last;    if(last != null) last.right = root;    last = root;    if(root.right != null) ConvertToList(root.right);}

---

实现二叉查找树的iterator

https://leetcode.com/problems/binary-search-tree-iterator/

/*** Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root* node of a BST.* Calling next() will return the next smallest number in the BST.* Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of * the tree.**/

思路：非递归中序遍历的实现，内置一个辅助栈，每次从栈顶取得元素，判断时判断栈顶是否为空即可。
由于中序遍历将所有元素遍历一遍，即将所有元素会入栈一次，而每次获取时会弹栈，即所有元素会弹出一次，因此调用n此next()函数的总时间复杂度为O(2n)，平均下来为O(2)是常数复杂度的，符合条件

public class BSTIterator {    Deque<TreeNode> deque = null;    public BSTIterator(TreeNode root) {        deque = new ArrayDeque<TreeNode>();        for(TreeNode head = root; head != null; deque.add(head), head = head.left);    }    /** @return whether we have a next smallest number */    public boolean hasNext() {        return deque.isEmpty() == true ? false : true;    }    /** @return the next smallest number */    public int next() {        TreeNode top = deque.pollLast();        for(TreeNode head = top.right; head != null; deque.add(head), head = head.left);        return top.val;    }}

---

f = (最大值+最小值)/2,设计一个算法，找出距离f值最近，大于f值得节点。

public TreeNode getNode(TreeNode root){    if(root == null) return null;    int max = 0, min = 0, mid = 0;    for(TreeNode tmp = root; tmp.right != null; max = tmp.val, tmp = tmp.right);    for(TreeNode tmp = root; tmp.left != null; min = tmp.val, tmp = tmp.left);    mid = (max + min)/2;    TreeNode rs = root, tmp = root;    while(tmp != null){        if(tmp.val < mid) tmp = tmp.right;        else{            rs = tmp;            tmp = tmp.left;        }    }    return rs;}