230. Kth Smallest Element in a BST

[Tcdian]

230. Kth Smallest Element in a BST

给定一个二叉搜索树，编写一个函数 kthSmallest 来查找其中第 k 个 最小的元素。

你可以假设 k 总是有效的，1 ≤ k ≤ 二叉搜索树元素个数。

Example 1

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1

Example 2

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3

[Tcdian]

Solution 1 ( 递归 )

• JavaScript Solution

/**
 * 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) {
    let result;
    inorderTravel(root);
    return result;
    function inorderTravel(root) {
        if (result !== undefined || root === null) {
            return;
        }
        inorderTravel(root.left);
        if (--k === 0) {
            result = root.val;
            return;
        }
        inorderTravel(root.right);
    }
};

Solution 2 ( 迭代 )

• JavaScript Solution

/**
 * 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) {
    let result;
    let patrol = root;
    const stack = [];
    while (stack.length !== 0 || patrol !== null) {
        while (patrol !== null) {
            stack.push(patrol);
            patrol = patrol.left;
        }
        patrol = stack.pop();
        if (--k === 0) {
            result = patrol.val;
            break;
        }
        patrol = patrol.right;
    }
    return result;
};