Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in sand all of this node‘s descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    /    4   5
  /  1   2

Given tree t:

   4 
  /  1   2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

     3
    /    4   5
  /  1   2
    /
   0

Given tree t:

   4
  /  1   2
題目描述：判斷t是不是s的子樹

方法一：遞歸
時間復雜度：o（n）             空間復雜度：o（1）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 
 
*/
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if(s==null) return false;
　　　　　// 考慮幾種情況，第一種s=t，調用isSubtreeStartRoot(s, t)返回true。第二種情況在根的左子樹或右子樹上，這需要遞歸了，因為不知道具體從哪個點開始。
        return isSubtreeStartRoot(s, t)||isSubtree(s.left,t)||isSubtree(s.right,t); 
    }
    
 
private boolean isSubtreeStartRoot(TreeNode s, TreeNode t){
        if(s==null&&t==null) return true;  //兩棵樹都走完了，說明兩棵樹一樣，返回true
        if(s==null||t==null) return false;  //一顆樹走完了，另一顆沒走完，那說明這兩個不一樣，返回false。
        if(s.val!=t.val) return false;
        return isSubtreeStartRoot(s.left,t.left)&&isSubtreeStartRoot(s.right,t.right);
    }
}

572. Subtree of Another Tree（子樹）