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Convert Sorted Array to Binary Search Tree

时间:2015-03-06 18:28:57      阅读:223      评论:0      收藏:0      [点我收藏+]

Convert Sorted Array to Binary Search Tree

问题:

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

思路:

  递归,dfs

我的代码:

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/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedArrayToBST(int[] num) {
        if(num == null || num.length == 0 )
            return null ;
        int len = num.length ;
        TreeNode rst = arrayToBST(num, 0, len - 1);
        return rst;
    }
    public TreeNode arrayToBST(int [] num, int left, int right)
    {
        if(left > right)    return null;
        if(left == right)   return new TreeNode(num[left]);
        int mid = (left + right)/2;
        TreeNode root = new TreeNode(num[mid]);
        root.left = arrayToBST(num, left, mid - 1);
        root.right = arrayToBST(num, mid + 1 , right);
        return root;
    }
}
View Code

他人代码:

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public TreeNode sortedArrayToBST(int[] num) {
    if (num.length == 0) {
        return null;
    }
    TreeNode head = helper(num, 0, num.length - 1);
    return head;
}

public TreeNode helper(int[] num, int low, int high) {
    if (low > high) { // Done
        return null;
    }
    int mid = (low + high) / 2;
    TreeNode node = new TreeNode(num[mid]);
    node.left = helper(num, low, mid - 1);
    node.right = helper(num, mid + 1, high);
    return node;
}
View Code

学习之处:

  • 我的代码有冗余的地方 left == right的判断可以省去

Convert Sorted Array to Binary Search Tree

原文:http://www.cnblogs.com/sunshisonghit/p/4318681.html

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