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LeetCode 98.Validate Binary Search Tree

2018-01-30 13:15 411 查看

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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标签:  二叉搜索树判断