LeetCode 572. Subtree of Another Tree

原题链接在这里：https://leetcode.com/problems/subtree-of-another-tree/#/description

题目：

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 s and 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

Return false.

题解：

Traverse s 的每个节点，看把该节点当成root的subtree是否与t identical.

如果s 和 t本身都是null, 那么t也算s的subtree, 因为null 是本身null 的subtree.

Time Complexity: O(m*n). m是s的节点数. n是t的节点数. 对于s的每一个节点都做了一次以该节点为root的subtree 与 t的比较.

Space: O(logm). stack space, 一直找不到的时候会走到s的底部.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSubtree(TreeNode s, TreeNode t) {
12         if(isSame(s, t)){
13             return true;
14         }
15         return s!=null && (isSubtree(s.left, t) || isSubtree(s.right,t ));
16     }
17     
18     private boolean isSame(TreeNode s, TreeNode t){
19         if(s == null && t == null){
20             return true;
21         }
22         
23         if(s == null || t == null){
24             return false;
25         }
26         
27         if(s.val != t.val){
28             return false;
29         }
30         
31         return isSame(s.left, t.left) && isSame(s.right, t.right);
32     }
33 }

对s 和 t 分别做preorder traversal, 然后看t得到的string 是不是 s得到string的 substring.

遇到null TreeNode时要添加特别符号, 不然就不能区分 123, null, null 和 1, 2, 3. 也无法区分 1, null, 2 和 1, 2, null. 这两种情况.

Time Complexity: O(m+n). m是s的节点数. n是t的节点数.

Space: O(Math.max(logm, logn)). Stack space.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSubtree(TreeNode s, TreeNode t) {
12         String sPreorder = preorderTraversal(s);
13         String tPreorder = preorderTraversal(t);
14         
15         return sPreorder.contains(tPreorder);
16     }
17     
18     private String preorderTraversal(TreeNode root){
19         StringBuilder sb = new StringBuilder();
20         Stack<TreeNode> stk = new Stack<TreeNode>();
21         stk.push(root);
22         while(!stk.isEmpty()){
23             TreeNode tn = stk.pop();
24             if(tn == null){
25                 sb.append(",#");
26             }else{
27                 sb.append(","+tn.val);
28                 stk.push(tn.left);
29                 stk.push(tn.right);
30             }
31         }
32         return sb.toString();
33     }
34 }

类似Binary Tree Preorder Traversal.