108.Kth Smallest Element in a BST

Given a binary search tree, write a function

kthSmallest

to find the kth
smallest element in it.

Note:

You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:

What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

分析：即找中序遇到的第k个结点。

/**
* 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。
* 中序遇到的第k个结点。
* @date 20160410
*/
public int kthSmallest(TreeNode root, int k) {
if(root == null){
return 0;
}else{
TreeNode node = root;
Stack stack = new Stack();
stack.push(node);
/*node表示入栈的元素，popnode表示弹栈的元素*/
while(!stack.isEmpty()){
while(node.left != null){//向左走到尽头，把最左分支的结点都入栈。
node = node.left;
stack.push(node);
}
TreeNode popnode= (TreeNode) stack.pop();//弹出栈顶元素
k--;
if(k == 0){
return popnode.val;
}
/*如果弹栈的元素有右孩子，则让其右孩子入栈，进行下一次循环*/
if(popnode.right != null){
node = popnode.right;
stack.push(node);
}
}
}
return 0;

}