leetcode 94: Valid Binary Search Tree

验证二叉搜索树

最新推荐文章于 2025-09-12 18:08:49 发布

转载最新推荐文章于 2025-09-12 18:08:49 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/xishibean/archive/2013/03/01/2951295.html

文章标签：

#java

本文提供了一种递归和非递归的方法来判断给定的二叉树是否为有效的二叉搜索树。有效二叉搜索树需满足左子树的所有节点小于根节点，右子树的所有节点大于根节点，且左右子树也必须是有效的二叉搜索树。

Validate Binary Search Tree Aug 31 '12

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

recursive approach:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 //null, root, {1,2,3} {2,3,1} {1,1} {1,#,1}
 
public class Solution {
    public boolean isValidBST(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        
        return valRec( root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
    
    private boolean valRec(TreeNode root, int low, int high) {
        if(root==null) return true;
        int x = root.val;
        if( x<=low || x>=high ) return false;
        
        return valRec(root.left, low, x) && valRec(root.right, x, high);  
    }
}

non-recursive approach:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 //null, root, {1,2,3} {2,3,1} {1,1} {1,#,1}
 
public class Solution {
    public boolean isValidBST(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root==null) return true;
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<Pair> pairStack = new Stack<Pair>();
        
        stack.push( root);
        pairStack.push( new Pair(Integer.MIN_VALUE, Integer.MAX_VALUE) );
        
        while(!stack.isEmpty() ){
            TreeNode n = stack.pop();
            Pair pair = pairStack.pop();
            int x = n.val;
            if( x<=pair.low || x>=pair.high) return false;
            if( n.left!=null) {
                stack.push( n.left);
                pairStack.push( new Pair(pair.low, x) );
            }
            
            if( n.right!=null) {
                stack.push( n.right);
                pairStack.push( new Pair(x, pair.high) );
            }
        }
        
        return true;
    }
    
    class Pair{
        int low;
        int high;
        Pair(int low, int high) {
            this.low=low;
            this.high=high;
        }
    }
}

转载于:https://www.cnblogs.com/xishibean/archive/2013/03/01/2951295.html