【LeetCode】 ValidateBinarySearchTree
2018-03-01 10:07
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/** * 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. * * */即:判断一个二叉树是否为二分查找树。何为二分查找树?二叉查找树(Binary Search Tree),也称有序二叉树(ordered binary tree),排序二叉树(sorted binary tree),是指一棵空树或者具有下列性质的二叉树:若任意节点的左子树不空,则左子树上所有结点的值均小于它的根结点的值;
任意节点的右子树不空,则右子树上所有结点的值均大于它的根结点的值;
任意节点的左、右子树也分别为二叉查找树。
没有键值相等的节点(no duplicate nodes)。
解题思路:1)中序遍历并存到List中。2)判断List是否为递增。(何为中序遍历? 看上篇文章:二叉树的遍历)
题解优点:思路简单,不易出错。
题解缺点:1、递归遍历,当二叉树太深时,会出现栈溢出。
2、Tree的节点个数未知,List需要自增,当Tree深度过大时,List频繁自增,重新分配内存,影响效率。
/**
* Created by xxxx on 2018/3/1.
*/
public class ValidateBinarySearchTreeImpl implements ValidateBinarySearchTree {
List<Integer> treeList = new ArrayList<Integer>();
@Override
public boolean isValidBST(TreeNode root) {
if (root == null) return false;
if (root.left == null && root.right == null) return false;
orderTree(root);
for (int i = 1; i <treeList.size(); i++){
if(treeList.get(i) <= treeList.get(i-1)){
return false;
}
}
return true;
}
private void orderTree(TreeNode root){
if(root!=null){
orderTree(root.left);
treeList.add(root.val);
orderTree(root.right);
}
}
}
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