标签：dfs null root arch color tree nod define span
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
Example 1:
2 / 1 3
Binary tree [2,1,3], return true.
Example 2:
1 / 2 3
Binary tree [1,2,3], return false.
判断一棵树是否是二叉搜索树。
判定范围即可。主要出问题的是出现在int的最大值，最小值附近。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { int flag1 = 0; int flag2 = 0; public boolean isValidBST(TreeNode root) { return isBST(root,Integer.MAX_VALUE,Integer.MIN_VALUE); } public boolean isBST(TreeNode root,int max,int min){ if( root == null ) return true; if( root.val == Integer.MAX_VALUE && max == root.val ){ if( flag1 == 0){ flag1 = 1; return isBST(root.right,max,root.val) && isBST(root.left,root.val,min); } else return false; } if( root.val == Integer.MIN_VALUE && min == root.val ){ if( flag2 == 0){ flag2 = 1; return isBST(root.right,max,root.val) && isBST(root.left,root.val,min); } else return false; } if( root.val <=min || root.val >= max) return false; return isBST(root.right,max,root.val) && isBST(root.left,root.val,min); } }
所以可以稍微优化一下。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isValidBST(TreeNode root) { return dfs(root, (long)(Integer.MIN_VALUE)-1, (long)(Integer.MAX_VALUE)+1); } private boolean dfs(TreeNode root, long gt, long lt){ if(root == null) return true; if(root.val >= lt || root.val <= gt) return false; return dfs(root.left, gt, root.val) && dfs(root.right, root.val, lt); } }
leetcode 98 Validate Binary Search Tree ----- java
标签：dfs null root arch color tree nod define span
原文地址：http://www.cnblogs.com/xiaoba1203/p/5997741.html
