Binary Search Tree Problems

Unique Binary Search Trees II

• class Solution {
      public List<TreeNode> generateTrees(int n) {
          return get(1, n);
      }
  
      private List<TreeNode> get(int left, int right) {
          List<TreeNode> l = new ArrayList<>();
          if (left > right) {
              l.add(null);
              return l;
          }
          for (int i = left; i <= right; i++) {
              List<TreeNode> leftList = get(left, i - 1);
              List<TreeNode> rightList = get(i + 1, right);
              for (TreeNode leftNode : leftList) {
                  for (TreeNode rightNode : rightList) {
                      TreeNode root = new TreeNode(i);
                      root.left = leftNode;
                      root.right = rightNode;
                      l.add(root);
                  }
              }
          }
          return l;
      }
  }

Unique Binary Search Trees

• class Solution {
      private Map<Integer, Integer> result = new HashMap<>();
  
      public int numTrees(int n) {
          if (result.containsKey(n))
              return result.get(n);
  
          if (n == 0 || n == 1)
              return 1;
  
          int sum = 0;
          for (int i = 1; i <= n; i++) {
              sum += numTrees(i - 1) * numTrees(n - i);
          }
  
          result.put(n, sum);
          return sum;
      }
  }

Validate Binary Search Tree

• 遍历

  class Solution {
      private boolean isBST = true;
      private Integer prev = null;
  
      public boolean isValidBST(TreeNode root) {
          traverse(root);
  
          return isBST;
      }
  
      private void traverse(TreeNode root) {
          if (root == null || !isBST)
              return;
  
          traverse(root.left);
          if (prev == null || root.val > prev)
              prev = root.val;
          else
              isBST = false;
          traverse(root.right);
      }
  }

• 分解

  class Solution {
      public boolean isValidBST(TreeNode root) {
          return _isValidBST(root, null, null);
      }
  
      private boolean _isValidBST(TreeNode root, TreeNode min, TreeNode max) {
          // min, max is the left and right border nearest to the current node
          if (root == null)
              return true;
          if (min != null && root.val <= min.val)
              return false;
          if (max != null && root.val >= max.val)
              return false;
  
          return _isValidBST(root.left, min, root) &&
                  _isValidBST(root.right, root, max);
      }
  }

Convert Sorted Array to Binary Search Tree

• class Solution {
      public TreeNode sortedArrayToBST(int[] nums) {
          return arrToBST(nums, 0, nums.length - 1);
      }
  
      private TreeNode arrToBST(int[] nums, int l, int r) {
          if (l > r)
              return null;
  
          if (l == r)
              return new TreeNode(nums[l]);
  
          int mid = l + (r - l) / 2;
          TreeNode root = new TreeNode(nums[mid]);
          root.left = arrToBST(nums, l, mid - 1);
          root.right = arrToBST(nums, mid + 1, r);
  
          return root;
      }
  }

Binary Search Tree Iterator

• class BSTIterator {
      private Deque<TreeNode> stack;
  
      public BSTIterator(TreeNode root) {
          stack = new ArrayDeque<>();
          while (root != null) {
              stack.push(root);
              root = root.left;
          }
      }
  
      // Time: amortized O(1)   Space: O(logN)
      public int next() {
          TreeNode n = stack.pop();
          int next = n.val;
  
          n = n.right;
          while (n != null) {
              stack.push(n);
              n = n.left;
          }
  
          return next;
      }
  
      public boolean hasNext() {
          return !stack.isEmpty();
      }
  
  }
  /**
   * Your BSTIterator object will be instantiated and called as such:
   * BSTIterator obj = new BSTIterator(root);
   * int param_1 = obj.next();
   * boolean param_2 = obj.hasNext();
   */

Kth Smallest Element in a BST

• class Solution {
      private int count;
      private int result;
  
      public int kthSmallest(TreeNode root, int k) {
          count = k;
          traverse(root);
  
          return result;
      }
  
      private void traverse(TreeNode root) {
          if (root == null || count == 0)
              return;
  
          traverse(root.left);
          if (count == 1)     result = root.val;
          count--;
          traverse(root.right);
      }
  }

• Follow up solution

  • Every TreeNode maintain another information size which store the size of current subtree whose root node is current node

Delete Node in a BST

• class Solution {
      public TreeNode deleteNode(TreeNode root, int key) {
          if (root == null)
              return null;
  
          if (root.val > key)
              root.left = deleteNode(root.left, key);
          else if (root.val < key)
              root.right = deleteNode(root.right, key);
          else {
              if (isLeaf(root))
                  root = null;
              else if (root.left == null)
                  root = root.right;
              else if (root.right == null)
                  root = root.left;
              else {
                  TreeNode leftMax = getMax(root.left);
                  leftMax.left = deleteNode(root.left, leftMax.val);
                  leftMax.right = root.right;
  
                  root = leftMax;
              }
          }
  
          return root;
      }
  
      private TreeNode getMax(TreeNode root) {
          while (root.right != null)
              root = root.right;
  
          return root;
      }
  
      private boolean isLeaf(TreeNode root) {
          return root.left == null && root.right == null;
      }
  }

Minimum Absolute Difference in BST

• class Solution {
      private int prev = -100000;
      private int minDiff = Integer.MAX_VALUE;
      public int getMinimumDifference(TreeNode root) {
          traverse(root);
  
          return minDiff;
      }
  
      private void traverse(TreeNode root) {
          if (root == null)
              return;
  
          traverse(root.left);
          minDiff = Math.min(minDiff, root.val - prev);
          prev = root.val;
          traverse(root.right);
      }
  }

Convert BST to Greater Tree

• class Solution {
      private int sum = 0;
  
      public TreeNode convertBST(TreeNode root) {
          traverse(root);
  
          return root;
      }
  
      public void traverse(TreeNode root) {
          if (root == null)
              return;
  
          traverse(root.right);
          sum += root.val;
          root.val = sum;
          traverse(root.left);
      }
  }

Search in a Binary Search Tree

• class Solution {
      public TreeNode searchBST(TreeNode root, int val) {
          if (root == null)
              return null;
          if (root.val == val)
              return root;
          else if (root.val < val)
              return searchBST(root.right, val);
          else
              return searchBST(root.left, val);
      }
  }

Insert into a Binary Search Tree

• class Solution {
      public TreeNode insertIntoBST(TreeNode root, int val) {
          if (root == null)
              return new TreeNode(val);
  
          if (val > root.val)
              root.right = insertIntoBST(root.right, val);
          else
              root.left = insertIntoBST(root.left, val);
  
          return root;
      }
  }