Subtree
最后更新于：2022-04-02 01:11:36
# 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)
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