98.Validate-Binary-Search-Tree

98. Validate Binary Search Tree

题目地址

https://leetcode.com/problems/validate-binary-search-tree/

http://www.lintcode.com/problem/validate-binary-search-tree/

http://www.jiuzhang.com/solutions/validate-binary-search-tree/

题目描述

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.


Example 1:

    2
   / \
  1   3

Input: [2,1,3]
Output: true

代码

Approach #1 Divide & Conquer

1. 传递upper和lower, 和当前节点进行比较

Complexity Analysis:

• Time complexis: O(n)

• Space complexity: O(N)

public class Solution {
    public boolean isValidBST(TreeNode root) {
    if (root == null) return true;

    return helper(root, Long.MIN_VALUE, Long.MAX_VALUE);
  }

  private boolean helper(TreeNode root, long lower, long upper) {
    if (root == null) return true;

    if (root.val >= upper || root.val <= lower)    return false;

    boolean isLeftValidBST = helper(root.left, lower, root.val);
    boolean isRightValidBST = helper(root.right, root.val, upper);

    return isLeftValidBST && isRightValidBST;
  }

}

Approach #2 Iteration

class Solution {
  LinkedList<TreeNode> stack = new LinkedList();
  LinkedList<TreeNode> uppers = new LinkedList();
  LinkedList<TreeNode> lowers = new LinkedList();

  public boolean isValidBST(TreeNode root) {
    Integer lower = null, upper = null, val;
    update(root, lower, upper);

    while (!stack.isEmpty()) {
      root = stack.poll();
      lower = lowers.poll();
      upper = uppers.poll();

      if (root == null) continue;
      val = root.val;
      if (lower != null && val <= lower) return false;
      if (upper != null && val >= upper) return false;
      update(root.right, val, upper);
      update(root.left, lower, val);
    }

    return true;
  }

     public void update(TreeNode root, Integer lower, Integer upper) {
    stack.add(root);
    lowers.add(lower);
    uppers.add(upper);
  }
}

Approach #3 No-Recursion, Access nodes from small to large

class Solution {
  public boolean isValidBST(TreeNode root) {
    Stack<TreeNode> stack = new Stack<>();

    while (root != null) {
      stack.push(root);
      root = root.left;
    }

    TreeNode lastNode = null;
    while (!stack.isEmpty()) {
      TreeNode node = stack.peek();
      if (lastNode != null && lastNode.val >= node.val) {
        return false;
      }
      lastNode = node;

      if (node.right == null) {
        node = stack.pop();
        while (!stack.isEmpty() && stack.peek().right == node) {
          node = stack.pop();
        }
      } else {
        node = node.right;
        while (node != null) {
          stack.push(node);
          node = node.left;
        }
      }
    }

    return false;
  }
}