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LeetCode (65):Same tree

时间:2015-10-03 21:58:55      阅读:164      评论:0      收藏:0      [点我收藏+]

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Total Accepted: 83663 Total Submissions: 200541 Difficulty: Easy

 Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

 

首先想到的是递归法,判断过程为依次遍历每一个点如果有:

  1. 如果都为null或是都不为null且值相等返回true
  2. 一个为Null另一个不为Null返回false
  3. 两个都不为null但值不相等返回false

代码如下:

class TreeNode{
       int val;
            TreeNode left;
           TreeNode right;
            TreeNode(int x) { val = x; }
}
public class Solution {
     public static boolean isSameTree(TreeNode p, TreeNode q) {
         if(p==null&&q==null)     return true;  //同时到达叶子节点
         else  if(p==null||q==null) return false;   //不同时到达叶子则不相同
         if(p.val!=q.val) return false;  //节点值不同则不相同
         else return isSameTree(p.left, q.left)&&isSameTree(p.right, q.right); //递归
                                                                    
     }
           //测试
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        TreeNode t1=new TreeNode(1);
        TreeNode nodeB=new TreeNode(2);
        t1.left=nodeB;
        
        TreeNode t2=new TreeNode(1);
        TreeNode nodeB1=new TreeNode(2);
        t2.left=nodeB1;        
        System.out.println(isSameTree(t1,t2));
    }

}

 

LeetCode (65):Same tree

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原文地址:http://www.cnblogs.com/rever/p/4853823.html

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