LeetCode#100. Same Tree
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- 题目:给定两个树,判断这棵树是不是一模一样(结构以及节点的值都需要一样)
- 难度:Easy
- 思路:直接采用深度优先对树进行递归,判断节点的值是否相等,加上对空指针的判断(结构上判断)。对于树的一些题目,很多时候可以用递归来做
- 代码:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { return recursion(p,q); } private static boolean recursion(TreeNode p, TreeNode q){ if(p == null && q != null){ return false; } if(p != null && q == null){ return false; } if(p == null && q == null){ return true; } if(p.val != q.val){ return false; }else{ if(!recursion(p.left, q.left)){ return false; }else{ return recursion(p.right, q.right); } } }}
- 更简洁的代码(思路一样,将两个指针都为空的情况先判断,则可以减少代码行数)
public boolean isSameTree(TreeNode p, TreeNode q) { if(p == null && q == null) return true; if(p == null || q == null) return false; if(p.val == q.val) return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); return false;}
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