Question

98. Validate Binary Search Tree

中文文档

Description

Given the root of a binary tree, _determine if it is a valid binary search tree (BST)_.

A valid 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:

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

Example 2:

Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

Solutions

Solution 1: Recursion

In-order traversal. If it is a valid binary search tree, then the sequence traversed should be monotonically increasing. So, we only need to compare and judge whether the current number traversed is greater than the previous number.

Alternatively, consider the subtree with root as the root, whether all node values are within the valid range, and judge recursively.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the tree.

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def isValidBST(self, root: Optional[TreeNode]) -> bool:
    def dfs(root):
      nonlocal prev
      if root is None:
        return True
      if not dfs(root.left):
        return False
      if prev >= root.val:
        return False
      prev = root.val
      if not dfs(root.right):
        return False
      return True

    prev = -inf
    return dfs(root)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private Integer prev;

  public boolean isValidBST(TreeNode root) {
    prev = null;
    return dfs(root);
  }

  private boolean dfs(TreeNode root) {
    if (root == null) {
      return true;
    }
    if (!dfs(root.left)) {
      return false;
    }
    if (prev != null && prev >= root.val) {
      return false;
    }
    prev = root.val;
    if (!dfs(root.right)) {
      return false;
    }
    return true;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  TreeNode* prev;

  bool isValidBST(TreeNode* root) {
    prev = nullptr;
    return dfs(root);
  }

  bool dfs(TreeNode* root) {
    if (!root) return true;
    if (!dfs(root->left)) return false;
    if (prev && prev->val >= root->val) return false;
    prev = root;
    if (!dfs(root->right)) return false;
    return true;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isValidBST(root *TreeNode) bool {
  var prev *TreeNode

  var dfs func(root *TreeNode) bool
  dfs = func(root *TreeNode) bool {
    if root == nil {
      return true
    }
    if !dfs(root.Left) {
      return false
    }
    if prev != nil && prev.Val >= root.Val {
      return false
    }
    prev = root
    if !dfs(root.Right) {
      return false
    }
    return true
  }

  return dfs(root)
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
  let prev = null;

  let dfs = function (root) {
    if (!root) {
      return true;
    }
    if (!dfs(root.left)) {
      return false;
    }
    if (prev && prev.val >= root.val) {
      return false;
    }
    prev = root;
    if (!dfs(root.right)) {
      return false;
    }
    return true;
  };

  return dfs(root);
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
public class Solution {
  private TreeNode prev;

  public bool IsValidBST(TreeNode root) {
    prev = null;
    return dfs(root);
  }

  private bool dfs(TreeNode root) {
    if (root == null)
    {
      return true;
    }
    if (!dfs(root.left))
    {
      return false;
    }
    if (prev != null && prev.val >= root.val)
    {
      return false;
    }
    prev = root;
    if (!dfs(root.right))
    {
      return false;
    }
    return true;
  }
}

Solution 2

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def isValidBST(self, root: Optional[TreeNode]) -> bool:
    def dfs(root, l, r):
      if root is None:
        return True
      if root.val <= l or root.val >= r:
        return False
      return dfs(root.left, l, root.val) and dfs(root.right, root.val, r)

    return dfs(root, -inf, inf)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public boolean isValidBST(TreeNode root) {
    return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE);
  }

  private boolean dfs(TreeNode root, long l, long r) {
    if (root == null) {
      return true;
    }
    if (root.val <= l || root.val >= r) {
      return false;
    }
    return dfs(root.left, l, root.val) && dfs(root.right, root.val, r);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  bool isValidBST(TreeNode* root) {
    return dfs(root, LONG_MIN, LONG_MAX);
  }

  bool dfs(TreeNode* root, long long l, long long r) {
    if (!root) return true;
    if (root->val <= l || root->val >= r) return false;
    return dfs(root->left, l, root->val) && dfs(root->right, root->val, r);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isValidBST(root *TreeNode) bool {
  return dfs(root, math.MinInt64, math.MaxInt64)
}

func dfs(root *TreeNode, l, r int64) bool {
  if root == nil {
    return true
  }
  v := int64(root.Val)
  if v <= l || v >= r {
    return false
  }
  return dfs(root.Left, l, v) && dfs(root.Right, v, r)
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function (root) {
  function dfs(root, l, r) {
    if (!root) {
      return true;
    }
    if (root.val <= l || root.val >= r) {
      return false;
    }
    return dfs(root.left, l, root.val) && dfs(root.right, root.val, r);
  }
  return dfs(root, -Infinity, Infinity);
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
public class Solution {
  public bool IsValidBST(TreeNode root) {
    return dfs(root, long.MinValue, long.MaxValue);
  }

  public bool dfs(TreeNode root, long l, long r) {
    if (root == null) {
      return true;
    }
    if (root.val <= l || root.val >= r) {
      return false;
    }
    return dfs(root.left, l, root.val) && dfs(root.right, root.val, r);
  }
}