[LeetCode]230. Kth Smallest Element in a BST

https://leetcode.com/problems/kth-smallest-element-in-a-bst/

找出BST中的第k个节点值

BST的中序遍历就是一个升序数组啊！！！！

数左子树中节点个数num，如果是k - 1个，那么当前root即为所求；如果num大于k，则从左子树中找第k个；否则从右子树中找第k - 1 - num个（即减去左子树中节点个数以及root这一个）

public class Solution {
public int kthSmallest(TreeNode root, int k) {
int num = count(root.left);
if (num >= k) {
return kthSmallest(root.left, k);
} else if (num + 1 < k) {
// 注意第二个参数
return kthSmallest(root.right, k - 1 - num);
}
return root.val;
}
private int count(TreeNode root) {
if (root == null) {
return 0;
}
return 1 + count(root.left) + count(root.right);
}
}

注意dfs的时候函数调用的时候就要保证root不是null，这样里面就不用判断了

递归二叉树的中序遍历

public class Solution {
int count = 0;
int num = 0;
public int kthSmallest(TreeNode root, int k) {
count = k;
dfs(root);
return num;
}
private void dfs(TreeNode root) {
if (root.left != null) {
dfs(root.left);
}
count--;
if (count == 0) {
num = root.val;
return;
}
if (root.right != null) {
dfs(root.right);
}
}
}

非递归二叉树的中序遍历

public class Solution {
public int kthSmallest(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();
k--;
if (k == 0) {
return cur.val;
}
cur = cur.right;
}
}
return -1;
}
}