230. 二叉搜索树中第K小的元素

Author: w2xi

README.md

@@ -83,6 +83,7 @@
 |222|[完全二叉树的节点个数](https://leetcode.cn/problems/count-complete-tree-nodes/)|[JavaScript](./algorithms/count-complete-tree-nodes.js)|Medium|
 |225|[用队列实现栈](https://leetcode.cn/problems/implement-stack-using-queues/)|[JavaScript](./algorithms/implement-stack-using-queues.js)|Easy|
 |226|[翻转二叉树](https://leetcode.cn/problems/invert-binary-tree/)|[JavaScript](./algorithms/invert-binary-tree.js)|Easy|
+|230|[二叉搜索树中第K小的元素](https://leetcode.cn/problems/kth-smallest-element-in-a-bst/)|[JavaScript](./algorithms/kth-smallest-element-in-a-bst.js)|Medium|
 |231|[2 的幂](https://leetcode.cn/problems/power-of-two/)|[JavaScript](./algorithms/power-of-two.js)|Easy|
 |232|[用栈实现队列](https://leetcode.cn/problems/implement-queue-using-stacks/)|[JavaScript](./algorithms/implement-queue-using-stacks.js)|Easy|
 |234|[回文链表](https://leetcode-cn.com/problems/palindrome-linked-list/)|[JavaScript](./algorithms/palindrome-linked-list.js)|Easy|

algorithms/kth-smallest-element-in-a-bst.js

@@ -0,0 +1,55 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} k
+ * @return {number}
+ */
+var kthSmallest = function (root, k) {
+  // 中序遍历二叉搜索树相当与遍历有序数组
+
+  // 递归 需要遍历整棵树
+  let counter = 0;
+  let rankKNum = 0;
+  const dfs = (root) => {
+    if (!root) return null;
+    dfs(root.left);
+    counter++;
+    if (k === counter) {
+      rankKNum = root.val;
+    }
+    dfs(root.right);
+  };
+  dfs(root);
+  return rankKNum;
+
+  // 迭代 无需遍历整棵树
+  // return inOrderIterate(root, k);
+};
+
+function inOrderIterate(root, k) {
+  const stack = [];
+  let curr = root,
+    node = null;
+  while (stack.length || curr) {
+    while (curr) {
+      stack.push(curr);
+      curr = curr.left;
+    }
+    node = stack.pop();
+    k--;
+    if (k === 0) {
+      break;
+    }
+    if (node.right) {
+      curr = node.right;
+    }
+  }
+  return node.val;
+}