# Subtree ### Source - lintcode: [(245) Subtree](http://www.lintcode.com/en/problem/subtree/#) ~~~ You have two every large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1. Example T2 is a subtree of T1 in the following case: 1 3 / \ / T1 = 2 3 T2 = 4 / 4 T2 isn't a subtree of T1 in the following case: 1 3 / \ \ T1 = 2 3 T2 = 4 / 4 Note A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical. ~~~ ### 题解 判断 T2是否是 T1的子树，首先应该在 T1中找到 T2的根节点，找到根节点后两棵子树必须完全相同。所以整个思路分为两步走：找根节点，判断两棵树是否全等。咋看起来极其简单，但实际实现中还是比较精妙的，尤其是递归的先后顺序及条件与条件或的处理。 ### Java ~~~ /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param T1, T2: The roots of binary tree. * @return: True if T2 is a subtree of T1, or false. */ public boolean isSubtree(TreeNode T1, TreeNode T2) { if (T2 == null) return true; if (T1 == null) return false; return identical(T1, T2) || isSubtree(T1.left, T2) || isSubtree(T1.right, T2); } private boolean identical(TreeNode T1, TreeNode T2) { if (T1 == null && T2 == null) return true; if (T1 == null || T2 == null) return false; if (T1.val != T2.val) return false; return identical(T1.left, T2.left) && identical(T1.right, T2.right); } } ~~~ ### 源码分析 这道题的异常处理相对 trick 一点，需要理解 null 对子树的含义。另外需要先调用`identical`再递归调用`isSubtree`判断左右子树的情况。方法`identical`中调用`.val`前需要判断是否为 null, 而后递归调用判断左右子树是否 identical。 ### 复杂度分析 identical 的调用，时间复杂度近似 O(n)O(n)O(n), 查根节点的时间复杂度随机，平均为 O(m)O(m)O(m), 故总的时间复杂度可近似为 O(mn)O(mn)O(mn). ### Reference - [LintCode: Subtree](http://cherylintcode.blogspot.com/2015/06/subtree.html)