# 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. ~~~ ### 題解 判斷 T2是否是 T1的子樹，首先應該在 T1中找到 T2的根節點，找到根節點后兩棵子樹必須完全相同。所以整個思路分為兩步走：找根節點，判斷兩棵樹是否全等。咋看起來極其簡單，但實際實現中還是比較精妙的，尤其是遞歸的先后順序及條件與條件或的處理。 ### 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); } } ~~~ ### 源碼分析 這道題的異常處理相對 trick 一點，需要理解 null 對子樹的含義。另外需要先調用`identical`再遞歸調用`isSubtree`判斷左右子樹的情況。方法`identical`中調用`.val`前需要判斷是否為 null, 而后遞歸調用判斷左右子樹是否 identical。 ### 復雜度分析 identical 的調用，時間復雜度近似 O(n)O(n)O(n), 查根節點的時間復雜度隨機，平均為 O(m)O(m)O(m), 故總的時間復雜度可近似為 O(mn)O(mn)O(mn). ### Reference - [LintCode: Subtree](http://cherylintcode.blogspot.com/2015/06/subtree.html)