题目:
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.
链接: http://leetcode.com/problems/same-tree/
题解:判断二叉树是否相同。 假如根节点都为空,则返回true; 根节点仅有一个为空,返回false; 根节点值相同,recursively比较两个子节点; 根节点值不同返回false。
Time Complexity - O(n), Space Complexity - O(n)
public class Solution { 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); else return false; } }
二刷:
先判断p和q为null的情况,接下来判断p和q的值是否相当,最后递归判断p和q左右子树是否相等。
也可以用stack或者queue来做两棵树的traversal。看了一下discuss,有用两个stack的,也有用一个stack的,以后再继续研究。
Java:
Time Complexity - O(n), Space Complexity - O(n)
/** * 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) { if (p == null || q == null) { return p == null && q == null; } if (p.val != q.val) { return false; } else { return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); } } }
三刷:
要好好学一下coding style的line wrapping
Java:
/** * 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) { if (p == null || q == null) { return p == q; } return (p.val == q.val) ? isSameTree(p.left, q.left) && isSameTree(p.right, q.right) : false; } }
Reference:
http://yohanan.org/steve/projects/java-code-conventions/#SECTION00052000000000000000
http://www.oracle.com/technetwork/articles/javase/codeconvtoc-136057.html
http://www.oracle.com/technetwork/java/codeconventions-150003.pdf