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.
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);
} }