LeetCode 98.Validate Binary Search Tree
2018-01-30 13:15
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LeetCode 98.Validate Binary Search Tree
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.
Example 1:
2 / \ 1 3
Binary tree
[2,1,3]
, return true.
Example 2:
1 / \ 2 3
Binary tree
[1,2,3]
, return false.
方法1:利用本身性质:左<根<右,初始时带入系统的最大值和最小值,在递归过程中换成它们自己的值,;
public boolean isValidBST(TreeNode root) { if(root==null) return true; return valid(root, Long.MIN_VALUE, Long.MAX_VALUE); } public boolean valid(TreeNode root,long low,long high){ if(root==null) return true; if(root.val<=low||root.val>=high) return false; return valid(root.left, low, root.val)&&valid(root.right, root.val, high); }
方法2:将二叉树进行中序遍历,若遍历结果是递增的有序序列,则是合法的二叉搜索树。
public boolean isValidBST(TreeNode root) { List<Integer> res=new ArrayList<>(); InOrderTree(res, root); for(int i=0;i<res.size()-1;i++){ if(res.get(i)>=res.get(i+1)) return false; } return true; } public void InOrderTree(List<Integer> list,TreeNode root){ if(root!=null){ InOrderTree(list, root.left); list.add(root.val); InOrderTree(list, root.right); } }
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