Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
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); } }
【leetcode】Validate Binary Search Tree（middle）
原文：http://www.cnblogs.com/dplearning/p/4481565.html