Validate Binary Search Tree

Question

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.
  • A single node tree is a BST

Example

An example:

  2
 / \
1   4
   / \
  3   5

The above binary tree is serialized as {2,1,4,#,#,3,5} (in level order).

题解1 - recursion

按照题中对二叉搜索树所给的定义递归判断,我们从递归的两个步骤出发分析:

  1. 基本条件/终止条件 - 返回值需斟酌。
  2. 递归步/条件递归 - 能使原始问题收敛。

终止条件好确定——当前节点为空,或者不符合二叉搜索树的定义,返回值分别是什么呢?先别急,分析下递归步试试先。递归步的核心步骤为比较当前节点的key和左右子节点的key大小,和定义不符则返回false, 否则递归处理。从这里可以看出在节点为空时应返回true, 由上层的其他条件判断。但需要注意的是这里不仅要考虑根节点与当前的左右子节点,还需要考虑左子树中父节点的最小值和右子树中父节点的最大值。否则程序在[10,5,15,#,#,6,20] 这种 case 误判。

由于不仅需要考虑当前父节点,还需要考虑父节点的父节点… 故递归时需要引入上界和下界值。画图分析可知对于左子树我们需要比较父节点中最小值,对于右子树则是父节点中的最大值。又由于满足二叉搜索树的定义时,左子结点的值一定小于根节点,右子节点的值一定大于根节点,故无需比较所有父节点的值,使用递推即可得上界与下界,这里的实现非常巧妙。

C++ - long long

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: True if the binary tree is BST, or false
*/
bool isValidBST(TreeNode *root) {
if (root == NULL) return true;

return helper(root, LLONG_MIN, LLONG_MAX);
}

bool helper(TreeNode *root, long long lower, long long upper) {
if (root == NULL) return true;

if (root->val <= lower || root->val >= upper) return false;
bool isLeftValidBST = helper(root->left, lower, root->val);
bool isRightValidBST = helper(root->right, root->val, upper);

return isLeftValidBST && isRightValidBST;
}
};

C++ - without long long

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param root: The root of binary tree.
* @return: True if the binary tree is BST, or false
*/
bool isValidBST(TreeNode *root) {
if (root == NULL) return true;

return helper(root, INT_MIN, INT_MAX);
}

bool helper(TreeNode *root, int lower, int upper) {
if (root == NULL) return true;

if (root->val <= lower || root->val >= upper) {
bool right_max = root->val == INT_MAX && root->right == NULL;
bool left_min = root->val == INT_MIN && root->left == NULL;
if (!(right_max || left_min)) {
return false;
}
}
bool isLeftValidBST = helper(root->left, lower, root->val);
bool isRightValidBST = helper(root->right, root->val, upper);

return isLeftValidBST && isRightValidBST;
}
};

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: True if the binary tree is BST, or false
*/
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;
// System.out.println("root.val = " + root.val + ", lower = " + lower + ", upper = " + upper);
// left node value < root node value < right node value
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;
}
}

源码分析

为避免节点中出现整型的最大最小值,引入 long 型进行比较。有些 BST 的定义允许左子结点的值与根节点相同,此时需要更改比较条件为root.val > upper. C++ 中 long 可能与 int 范围相同,故使用 long long. 如果不使用比 int 型更大的类型,那么就需要在相等时多加一些判断。

复杂度分析

递归遍历所有节点,时间复杂度为 $$O(n)$$, 使用了部分额外空间,空间复杂度为 $$O(1)$$.

题解2 - iteration

联想到二叉树的中序遍历。

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isValidBST(TreeNode root) {
Deque<TreeNode> st = new ArrayDeque<>();
long pre = Long.MIN_VALUE;
// inorder traverse
while (root != null || !st.isEmpty()) {
if (root != null) {
st.push(root);
root = root.left;
}
else {
root = st.pop();
if (root.val > pre)
pre = root.val;
else
return false;
root = root.right;
}
}
return true;
}
}

Reference




评论

Your browser is out-of-date!

Update your browser to view this website correctly. Update my browser now

×