Question

98. 验证二叉搜索树

English Version

题目描述

给你一个二叉树的根节点 root ，判断其是否是一个有效的二叉搜索树。

有效 二叉搜索树定义如下：

• 节点的左子树只包含 小于 当前节点的数。
• 节点的右子树只包含 大于 当前节点的数。
• 所有左子树和右子树自身必须也是二叉搜索树。

 

示例 1：

输入：root = [2,1,3]
输出：true

示例 2：

输入：root = [5,1,4,null,null,3,6]
输出：false
解释：根节点的值是 5 ，但是右子节点的值是 4 。

 

提示：

• 树中节点数目范围在[1, 104] 内
• -231 <= Node.val <= 231 - 1

解法

方法一：递归

中序遍历，若是一个有效的二叉搜索树，那么遍历到的序列应该是单调递增的。所以只要比较判断遍历到的当前数是否大于上一个数即可。

或者考虑以 root 为根的子树，所有节点的值是否都在合法范围内，递归判断即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是树中节点的数量。

# 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;
  }
}

方法二

# 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);
  }
}