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.
算法思路：
思路1：求出其中序遍历序列，判断其实是否为递增序列，空间复杂度为O(n)
思路2：二分法。每一个点的上下界都是确定的。
代码如下：
1 public class Solution { 2 public boolean isValidBST(TreeNode root) { 3 if(root == null || (root.left == null && root.right == null)) return true; 4 return isValid(root, Integer.MAX_VALUE, Integer.MIN_VALUE); 5 } 6 private boolean isValid(TreeNode root,int biggest,int smallest){ 7 if(root == null ) return true; 8 if(root.val >= biggest || root.val <= smallest) return false; 9 else 10 return isValid(root.left, root.val, smallest) && isValid(root.right, biggest, root.val); 11 } 12 }
[leetcode]Validate Binary Search Tree,布布扣,bubuko.com
[leetcode]Validate Binary Search Tree
原文：http://www.cnblogs.com/huntfor/p/3881559.html