lintcode: 二叉查找树中搜索区间

题目

二叉查找树中搜索区间 

给定两个值 k1 和 k2(k1 < k2)和一个二叉查找树的根节点。找到树中所有值在 k1 到 k2 范围内的节点。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找树的中的节点值。返回所有升序的节点值。

样例

如果有 k1 = 10 和 k2 = 22, 你的程序应该返回 [12, 20, 22].

    20
   /  \
  8   22
 / \
4   12

解题

已经是二叉排序树了,中序遍历的结果就是升序的,然后在升序的序列中找到符合条件的部分节点就好了

《lintcode: 二叉查找树中搜索区间》
《lintcode: 二叉查找树中搜索区间》

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of the binary search tree.
     * @param k1 and k2: range k1 to k2.
     * @return: Return all keys that k1<=key<=k2 in ascending order.
     */
    public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
        // write your code here
        ArrayList<Integer> inorder = inorderTraveral(root);
        ArrayList<Integer> result = new ArrayList<Integer>();
        for(int i=0;i< inorder.size();i++){
            int tmp = inorder.get(i);
            if(k1<= tmp && tmp <= k2){
                result.add(tmp);
            }
        }
        return result;
    }
    public ArrayList<Integer> inorderTraveral(TreeNode root){
        ArrayList<Integer> inorder = new ArrayList<Integer>();
        if(root == null){
            return inorder;
        }
        inorder.addAll(inorderTraveral(root.left));
        inorder.add(root.val);
        inorder.addAll(inorderTraveral(root.right));
        return inorder;
    }
}

Java Code

也可以直接中序遍历过程中,符合条件的节点值加入到ArrayList中去

《lintcode: 二叉查找树中搜索区间》
《lintcode: 二叉查找树中搜索区间》

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of the binary search tree.
     * @param k1 and k2: range k1 to k2.
     * @return: Return all keys that k1<=key<=k2 in ascending order.
     */
    public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
        // write your code here
        ArrayList<Integer> inorder = inorderTraveral(root,k1,k2);
        
        return inorder;
    }
    public ArrayList<Integer> inorderTraveral(TreeNode root,int k1,int k2){
        ArrayList<Integer> inorder = new ArrayList<Integer>();
        if(root == null){
            return inorder;
        }
        inorder.addAll(inorderTraveral(root.left,k1,k2));
        if( root.val >= k1 && root.val <= k2)
            inorder.add(root.val);
        inorder.addAll(inorderTraveral(root.right,k1,k2));
        return inorder;
    }
}

Java Code

Python 实现

Python 中 append 是在list 后面添加元素

+= 是连接两个list

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""
class Solution:
    """
    @param root: The root of the binary search tree.
    @param k1 and k2: range k1 to k2.
    @return: Return all keys that k1<=key<=k2 in ascending order.
    """     
    def searchRange(self, root, k1, k2):
        # write your code here
        return self.inorderTraveral(root,k1,k2)
        
    def inorderTraveral(self,root,k1,k2):
        inorder = list()
        if root == None:
            return inorder
        left = self.inorderTraveral(root.left,k1,k2)
        if len(left)!=0:
            inorder += left
        if root.val>= k1 and root.val <= k2:
            inorder.append(root.val)
        right = self.inorderTraveral(root.right,k1,k2)
        if len(right)!=0:
            inorder += right
        return inorder

 



    原文作者:水滴失船
    原文地址: https://www.cnblogs.com/theskulls/p/5097990.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞