标签：ret return bool 相等 cti check ini for function
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.
思路：
这道题要判断树形一样，在这个基础上val一样。
共有5种树型：
1. node为null
2. node是leave
3. node只有左child
4. node只有右child
5. node有两个children
Base cases:
1和2是base case树形
先判断是不是两个node都为null，如果是，return true;
在判断是不是node一个为null一个不为null，如果是，return false;
如果两个node都是leaves, 判断val是不是相等，等则true，不等则false;
Recursive cases:
3, 4和5是recursive case树形
树形均为3: 判断val相等和左child相等
树形均为4: 判断val相等和右child相等
树形均为5: 判断val相等, 左child相等和右child相等
其他：树形不一样
/** * 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) { // base case: // two nodes are both null: true // one is null and another is not: false // both are leaves: check the val of the two nodes if (p == null && q == null) { return true; } else if (p == null || q == null) { return false; } else if (p.left == null && q.left == null && p.right == null && q.right == null) { if (p.val == q.val) { return true; } else { return false; } // recursive case: } else if (p.left == null && q.left == null && p.right != null && q.right != null) { return p.val == q.val && isSameTree(p.right, q.right); } else if (p.left != null && q.left != null && p.right == null && q.right == null) { return p.val == q.val && isSameTree(p.left, q.left); } else if (p.left != null && q.left != null && p.right != null && q.right != null) { return (p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right)); } else { return false; } } }
标签：ret return bool 相等 cti check ini for function
原文地址：http://www.cnblogs.com/fqmai/p/6216643.html
