【leetcode】Validate Binary Search Tree（middle）

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.

思路：中序遍历。当前值要比之前的小。

bool isValidBST(TreeNode* root) {
        TreeNode * pPre = NULL;
        TreeNode * pCur = root;
        vector<TreeNode *> v;

        while(!v.empty() || NULL != pCur)
        {
            if(NULL != pCur)
            {
                v.push_back(pCur);
                pCur = pCur->left;
            }
            else
            {
                if(pPre != NULL && v.back()->val <= pPre->val)
                    return false;
                pPre = v.back();    
                v.pop_back();
                pCur = pPre->right;
            }
        }
        return true;
    }

大神递归版：注意，每次左子树的值范围在最小值和根值之间，右子树的范围在根植和最大值之间。

public 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);
    }
}