寻找二叉搜索树的第K小的节点

题目要求：

230. Kth Smallest Element in a BST

 

Question
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Difficulty: Medium

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?
Hint:
Try to utilize the property of a BST.
What if you could modify the BST node's structure?
The optimal runtime complexity is O(height of BST).
Credits:

Special thanks to @ts for adding this problem and creating all test cases.

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三种解法：
public int kthSmallest(TreeNode root, int k) {
int count=countNodes(root.left)+1;
if(k==count) return root.val;
if(k<count) return kthSmallest(root.left,k);
else{
return kthSmallest(root.right,k-count);
}
}
public int countNodes(TreeNode root){
if(root==null) return 0;
return 1+countNodes(root.left)+countNodes(root.right);
}以上解法最漂亮，用到了二叉树的结构特性与题目要求
private static int number = 0;
private static int count = 0;

public int kthSmallest(TreeNode root, int k) {
count = k;
helper(root);
return number;
}

public void helper(TreeNode n) {
if (n.left != null) helper(n.left);
count--;
if (count == 0) {
number = n.val;
return;
}
if (n.right != null) helper(n.right);
}上面的解法就是中序遍历的解法，count  与  number  都应设置喂全局变量
public int kthSmallest(TreeNode root, int k) {
Stack<TreeNode> st = new Stack<>();

while (root != null) {
st.push(root);
root = root.left;
}

while (k != 0) {
TreeNode n = st.pop();
k--;
if (k == 0) return n.val;
TreeNode right = n.right;
while (right != null) {
st.push(right);
right = right.left;
}
}

return -1; // never hit if k is valid
}第三种解法应用了一个栈，与使用队列的效果是一样的，但是在可理解性上不如应用队列。