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

 

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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:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. 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。