230. Kth Smallest Element in a BST - jiejackyzhang/leetcode-note GitHub Wiki

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).

Tree类题目。 BST的特点是如果inorder traversal可以得到排序的结果。 因此采用这种思路，第k个即为所求。

##Approach 1: iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        while(node != null) {
            stack.push(node);
            node = node.left;
        }
        while(!stack.empty()) {
            node = stack.pop();
            k--;
            if(k == 0)  return node.val;
            node = node.right;
            while(node != null) {
                stack.push(node);
                node = node.left;
            }
        }
        return -1;
    }
}

##Binary Search 每次计算一下左子树的节点个数count，

  • 如果k==count+1，则root即为所求；
  • 如果k<=count，则在左子树中继续寻找kth smallest element；
  • 如果k>count+1，则在右子树中继续寻找(k-count-1)th smallest element。
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int count = countNodes(root.left);
        if(k <= count) {
            return kthSmallest(root.left, k);
        } else if(k > count + 1) {
            return kthSmallest(root.right, k-count-1);
        }
        return root.val;
    }
    
    private int countNodes(TreeNode root) {
        if(root == null) return 0;
        return 1 + countNodes(root.left) + countNodes(root.right);
    }
}
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