标签：contain min pen only [] rmi sea nod with
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
Example 1:
Input:
2
/ 1 3
Output: true
Example 2:
5 / 1 4 / 3 6 Output: false Explanation: The input is: [5,1,4,null,null,3,6]. The root node‘s value is 5 but its right child‘s value is 4.
Approach #1: C++.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
vector<int> inorderTraverseArray;
inorder(root, inorderTraverseArray);
for (int i = 1; i < inorderTraverseArray.size(); ++i) {
if (inorderTraverseArray[i-1] >= inorderTraverseArray[i]) return false;
}
return true;
}
private:
void inorder(TreeNode* root, vector<int>& inorderTraverseArray) {
if (root == NULL) return ;
if (root->left != NULL) inorder(root->left, inorderTraverseArray);
inorderTraverseArray.push_back(root->val);
if (root->right != NULL) inorder(root->right, inorderTraverseArray);
}
};
Approach #2: Java.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
if (root == null) return true;
if (root.val >= maxVal || root.val <= minVal) return false;
return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
}
}
Approach #3: Python.
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
output = []
self.solve(root, output)
for i in range(1, len(output)):
if output[i-1] >= output[i]:
return False
return True
def solve(self, root, output):
if root == None:
return
self.solve(root.left, output)
output.append(root.val)
self.solve(root.right, output)
标签：contain min pen only [] rmi sea nod with
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10011104.html
