Subtree of Another Tree.java

E
1533279340
tags: Tree, DFS, Divide and Conquer

#### Tree 
- Traverse tree: left, right
- Concept of partial compare vs. whole compare

```
/*
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.
 */


 /**
 * 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 || t == null) return s == null && t == null;

        return sameTree(s, t) || isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    private boolean sameTree(TreeNode s, TreeNode t) {
        if (s == null || t == null) return s == null && t == null;
        
        return s.val == t.val && sameTree(s.left, t.left) && sameTree(s.right, t.right);
    }
}

/*
Thoughts: similar to compare identical trees.
Except: only start compare if s.val == t.val, otherwise, keep dfs.
*/
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if (s == null || t == null) return s == null && t == null;

        return (s.val == t.val && sameTree(s, t)) || isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    private boolean sameTree(TreeNode s, TreeNode t) {
        if (s == null || t == null) return s == null && t == null;
        
        return s.val == t.val && sameTree(s.left, t.left) && sameTree(s.right, t.right);
    }
}
```