【Leetcode】Valid Binary Search Tree
2015-07-17 23:26
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【题目】
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.
【思路】
查看clean book
【代码】
/**
* 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 valid(root, null,null);
}
public boolean valid(TreeNode root, Integer low, Integer high){
if(root == null) return true;
return (low == null || root.val > low) && (high == null || root.val < high) &&
valid(root.left,low,root.val) && valid(root.right,root.val,high);
}
}
DFS
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.
【思路】
查看clean book
【代码】
/**
* 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 valid(root, null,null);
}
public boolean valid(TreeNode root, Integer low, Integer high){
if(root == null) return true;
return (low == null || root.val > low) && (high == null || root.val < high) &&
valid(root.left,low,root.val) && valid(root.right,root.val,high);
}
}
DFS
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