https://leetcode.com/problems/validate-binary-search-tree/description/
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
Example 2:
5
/ \
1 4
/ \
3 6
Input: [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.
这道题是让你验证一棵树是否为二叉查找树（BST）。 由于中序遍历的性质如果一个树遍历的结果是有序数组，那么他也是一个二叉查找树(BST),
我们只需要中序遍历，然后两两判断是否有逆序的元素对即可，如果有，则不是BST，否则即为一个BST。
根据定义，一个结点若是在根的左子树上，那它应该小于根结点的值而大于左子树最大值；若是在根的右子树上，那它应该大于根结点的值而小于右子树最小值。也就是说，每一个结点必须落在某个取值范围：
- 根结点的取值范围为（考虑某个结点为最大或最小整数的情况）：(long_min, long_max)
- 左子树的取值范围为：(current_min, root.value)
- 右子树的取值范围为：(root.value, current_max)
- 二叉树的基本操作（遍历）
- 中序遍历一个二叉查找树（BST）的结果是一个有序数组
- 如果一个树遍历的结果是有序数组，那么他也是一个二叉查找树(BST)
- 语言支持：JS，C++, Java
JavaScript Code：
/*
* @lc app=leetcode id=98 lang=javascript
*
* [98] Validate Binary Search Tree
*/
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isValidBST = function(root) {
if (root === null) return true;
if (root.left === null && root.right === null) return true;
const stack = [root];
let cur = root;
let pre = null;
function insertAllLefts(cur) {
while(cur && cur.left) {
const l = cur.left;
stack.push(l);
cur = l;
}
}
insertAllLefts(cur);
while(cur = stack.pop()) {
if (pre && cur.val <= pre.val) return false;
const r = cur.right;
if (r) {
stack.push(r);
insertAllLefts(r);
}
pre = cur;
}
return true;
};C++ Code：
// 递归
class Solution {
public:
bool isValidBST(TreeNode* root) {
TreeNode* prev = nullptr;
return validateBstInorder(root, prev);
}
private:
bool validateBstInorder(TreeNode* root, TreeNode*& prev) {
if (root == nullptr) return true;
if (!validateBstInorder(root->left, prev)) return false;
if (prev != nullptr && prev->val >= root->val) return false;
prev = root;
return validateBstInorder(root->right, prev);
}
};
// 迭代
class Solution {
public:
bool isValidBST(TreeNode* root) {
auto s = vector<TreeNode*>();
TreeNode* prev = nullptr;
while (root != nullptr || !s.empty()) {
while (root != nullptr) {
s.push_back(root);
root = root->left;
}
root = s.back();
s.pop_back();
if (prev != nullptr && prev->val >= root->val) return false;
prev = root;
root = root->right;
}
return true;
}
};Java Implementation
/**
* 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) {
Stack<Integer> stack = new Stack<> ();
TreeNode previous = null;
while (root != null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if (previous != null && root.val <= previous.val ) return false;
previous = root;
root = root.right;
}
return true;
}
}- 语言支持：C++，Python3, Java
C++ Code：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// 递归
class Solution {
public:
bool isValidBST(TreeNode* root) {
return helper(root, LONG_MIN, LONG_MAX);
}
private:
bool helper(const TreeNode* root, long min, long max) {
if (root == nullptr) return true;
if (root->val >= max || root->val <= min) return false;
return helper(root->left, min, root->val) && helper(root->right, root->val, max);
}
};
// 循环
class Solution {
public:
bool isValidBST(TreeNode* root) {
if (root == nullptr) return true;
auto ranges = queue<pair<long, long>>();
ranges.push(make_pair(LONG_MIN, LONG_MAX));
auto nodes = queue<const TreeNode*>();
nodes.push(root);
while (!nodes.empty()) {
auto sz = nodes.size();
for (auto i = 0; i < sz; ++i) {
auto range = ranges.front();
ranges.pop();
auto n = nodes.front();
nodes.pop();
if (n->val >= range.second || n->val <= range.first) {
return false;
}
if (n->left != nullptr) {
ranges.push(make_pair(range.first, n->val));
nodes.push(n->left);
}
if (n->right != nullptr) {
ranges.push(make_pair(n->val, range.second));
nodes.push(n->right);
}
}
}
return true;
}
};Python Code:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isValidBST(self, root: TreeNode, area: tuple=(-float('inf'), float('inf'))) -> bool:
"""思路如上面的分析，用Python表达会非常直白
:param root TreeNode 节点
:param area tuple 取值区间
"""
if root is None:
return True
is_valid_left = root.left is None or\
(root.left.val < root.val and area[0] < root.left.val < area[1])
is_valid_right = root.right is None or\
(root.right.val > root.val and area[0] < root.right.val < area[1])
# 左右节点都符合，说明本节点符合要求
is_valid = is_valid_left and is_valid_right
# 递归下去
return is_valid\
and self.isValidBST(root.left, (area[0], root.val))\
and self.isValidBST(root.right, (root.val, area[1]))Java Code:
/**
* 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 helper(root, null, null);
}
private boolean helper(TreeNode root, Integer lower, Integer higher) {
if (root == null) return true;
if (lower != null && root.val <= lower) return false;
if (higher != null && root.val >= higher) return false;
if (!helper(root.left, lower, root.val)) return false;
if (!helper(root.right, root.val, higher)) return false;
return true;
}
}