230. Kth Smallest Element in a BST

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?

Solution1：二分查找

思路：在当前结点，数左子树的结点总数num，(1)如果不到k - 1，说明在右子树里面；(2)如果等于k-1，当前结点就是kthSmallest；(3)如果大于k-1,说明就在左子树里面，递归继续在左子树里面找(k-num-1)thSmallest，不用管右子树了。
时间复杂度: 如果一直是情况(3)，就是O(N)，如果一直是情况(1)必然是O(N^2)，

Average: 当是树是基本平衡的时候，T(N) = T(N / 2) + O(N/2) => O(N)
worst: O(N ^ 2) 比如只有左子树，target还在最左下
Time Complexity: O(N) Space Complexity: O(N)

Solution2.a：二分查找 + augmented TreeNode data structure

(不如2.b)
思路：将结点的结构多加一个lCount，记录sub_left_tree的结点总数。
这样在每次调用kthSmallest()时，不用每次再count，直接看lCount二分就能找到，是O(logN)。
但是这是在新树已经建立好的情况下，如果是按题目的话建立树的过程这里对整体树复制，因为用Iterative由上倒下count需要O(logN)，整体也还是需要O(NlogN)；如果是从的数组建立的话，排序O(NlogN) + 累积Count(O(N)) + 建树O(N)

2.b 思路与a大体相同，count_all_sub 和 count_left_sub是一样的。用分治法建树只需要O(N)，由低向上count

Solution3.a：In-order Traversal (recursive)

Time Complexity: O(k), worst case O(N) Space Complexity: O(N) 递归缓存

Solution3.b：In-order Traversal (iterative)

Time Complexity: O(k) worst case O(N) Space Complexity: O(N)

Solution1 Code：

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        
        int left_count = countSubTree(root.left);
        if(left_count == k - 1) {
            return root.val;
        }
        else if(left_count > k - 1) {
            // kthSmallest element in the left tree
            return kthSmallest(root.left, k);
        }
        else {
            // kthSmallest element in the right tree
            return kthSmallest(root.right, k - left_count - 1);
        }
    }
    
    private int countSubTree(TreeNode root) {
        if(root == null) return 0;
        return countSubTree(root.left) + countSubTree(root.right) + 1;
    }
}

Solution2 Code：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class TreeNodeWithCount {
    int val;
    int lCount;
    TreeNodeWithCount left;
    TreeNodeWithCount right;
    
    TreeNodeWithCount(int x) { val = x; }
}

public class Solution {

    public int kthSmallest(TreeNode root, int k) {
        if(root == null) return -1;
        TreeNodeWithCount rootWithCount = createBSTWithCount(root);
        return kthSmallestWithCount(rootWithCount, k);
    }

    public TreeNodeWithCount createBSTWithCount(TreeNode root) {
        Queue queue = new LinkedList();
        queue.offer(root);
        TreeNodeWithCount rootWithCount = null;
        while(!queue.isEmpty()) {
            TreeNode node = queue.poll();
            
            TreeNodeWithCount nodeWithCount = new TreeNodeWithCount(node.val);
            rootWithCount = insertBSTWithCount(rootWithCount, nodeWithCount);
            
            if(node.left != null) queue.add(node.left);
            if(node.right != null) queue.add(node.right);
        }
        return rootWithCount;
    }

    public TreeNodeWithCount insertBSTWithCount(TreeNodeWithCount rootWithCount, TreeNodeWithCount nodeWithCount) {
        TreeNodeWithCount cur = rootWithCount, parent = rootWithCount;
        while(cur != null) {
            parent = cur;
            if(nodeWithCount.val < cur.val) {
                cur.lCount++;
                cur = cur.left;
            } else {
                cur = cur.right;
            }
        }
        if(rootWithCount == null) {
            rootWithCount = nodeWithCount;
        } else if(nodeWithCount.val < parent.val) {
            parent.left = nodeWithCount;
        } else {
            parent.right = nodeWithCount;
        }
        return rootWithCount;
    }

    public int kthSmallestWithCount(TreeNodeWithCount rootWithCount, int k) {
        while(rootWithCount != null) {
            if(k == rootWithCount.lCount + 1) {
                return rootWithCount.val;
            } else if(k <= rootWithCount.lCount) {
                rootWithCount = rootWithCount.left;
            } else {
                k = k - rootWithCount.lCount - 1;
                rootWithCount = rootWithCount.right;
            }
        }
        return -1;
    }
}

Solution2.b Code：

   public class Solution {
        public int kthSmallest(TreeNode root, int k) {
            TreeNodeWithCount rootWithCount = buildTreeWithCount(root);
            return kthSmallest(rootWithCount, k);
        }
        
        private TreeNodeWithCount buildTreeWithCount(TreeNode root) {
            if (root == null) return null;
            TreeNodeWithCount rootWithCount = new TreeNodeWithCount(root.val);
            rootWithCount.left = buildTreeWithCount(root.left);
            rootWithCount.right = buildTreeWithCount(root.right);
            if (rootWithCount.left != null) rootWithCount.count += rootWithCount.left.count;
            if (rootWithCount.right != null) rootWithCount.count += rootWithCount.right.count;
            return rootWithCount;
        }
        
        private int kthSmallest(TreeNodeWithCount rootWithCount, int k) {
            if (k <= 0 || k > rootWithCount.count) return -1;
            if (rootWithCount.left != null) {
                if (rootWithCount.left.count >= k) return kthSmallest(rootWithCount.left, k);
                if (rootWithCount.left.count == k-1) return rootWithCount.val;
                return kthSmallest(rootWithCount.right, k-1-rootWithCount.left.count);
            } else {
                if (k == 1) return rootWithCount.val;
                return kthSmallest(rootWithCount.right, k-1);
            }
        }
        
        class TreeNodeWithCount {
            int val;
            int count;
            TreeNodeWithCount left;
            TreeNodeWithCount right;
            TreeNodeWithCount(int x) {val = x; count = 1;};
        }
    }

Solution3.a Code：

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

Solution3.a.2 Round1 Code：

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        TreeNode result = dfs(root, k);
        if(result == null) return 0;
        else return result.val;
    }
    
    private int count = 0;
    
    private TreeNode dfs(TreeNode root, int k) {
        if(root == null) return null;
        
        TreeNode l_res = dfs(root.left, k);
        if(l_res != null) return l_res;
        
        count++;
        if(count == k) return root;
        
        TreeNode r_res = dfs(root.right, k);
        if(r_res != null) return r_res;
        
        return null;
    }
}

Solution3.b Code：

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int count = k;
        TreeNode cur = root;
        Stack stack = new Stack<>();
        while(cur != null || !stack.empty()){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            count--;
            if(count == 0) return cur.val;
            
            cur = cur.right;
        }
        return Integer.MIN_VALUE; //not found
    }
}