【Binary tree】Verify binary search tree

0x00 topic

Given a binary tree
, determine whether it is a valid二叉搜索树

A binary search tree has the following 特征:
A node 左's subtree contains only 小于the current node's tree
A node 右's subtree contains only 大于the current node's tree All
subtrees and subtrees must also be左右二叉搜索树

---

0x01 Ideas

Binary search tree, 中序traversal, is an 升序array
to 当前compare 上一个the size of the value of the node with the value of the node, you can

---

0x02 solution

language:Swift

tree node:TreeNode

public class TreeNode {
    public var val: Int
    public var left: TreeNode?
    public var right: TreeNode?
    public init() { self.val = 0; self.left = nil; self.right = nil; }
    public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
    public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
        self.val = val
        self.left = left
        self.right = right
    }
}

solution:

func isValidBST(_ root: TreeNode?) -> Bool {
    // 记录前一个节点的值
    var pre = Int.min
    
    func dfs(_ root: TreeNode?) -> Bool {
        guard let root = root else {
            return true
        }
        
        // 前序遍历位置
        let left = dfs(root.left)
        
        // 中序遍历位置
        // 大于当前节点，则返回 false
        if pre >= root.val {
            return false
        }else{
            // 记录当前节点值
            pre = root.val
        }
        
        // 后序遍历位置
        let right = dfs(root.right)
        
        // 返回 左子树 与 右子树 的结果
        return left && right
    }
    
    return dfs(root)
}

---

small notes

---