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Leetcode 链表

时间:2018-12-02 14:09:15      阅读:180      评论:0      收藏:0      [点我收藏+]
  public ListNode removeNthFromEnd(ListNode head, int n) {
        int res = onePass(head, n);
        return res == n ? head.next : head;
    }
    
    int onePass(ListNode curr, int n) {
        if(curr == null) return 0;
        int res = onePass(curr.next, n);
        if(res == n) {
            curr.next = curr.next.next;
        }
        return res + 1;
    }
//转自leetcode

19. Remove Nth Node From End of List

 

minimum-depth-of-binary-tree

Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class Solution {
    public int find(TreeNode root) {
        int a1=99999,a2=99999;
        if(root.left==null&&root.right==null) //找到叶节点 
        {
            return 1;
        }
          if(root.left!=null)
          {
              a1=find(root.left);
          }
        if(root.right!=null)
        {
            a2=find(root.right);
        }
        return Math.min(a1,a2)+1;
    }
    public int run(TreeNode root) {
        if(root==null) return 0; //这句遗漏了,要考虑为空的情况
     int t= find(root);
       
        return t;
    }
}

 

Leetcode 链表

原文:https://www.cnblogs.com/LandingGuy/p/10053042.html

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