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# 572.Subtree-of-Another-Tree

## 572. Subtree of Another Tree

## 题目地址

<https://leetcode.com/problems/subtree-of-another-tree/>

## 题目描述

```
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.
```

## 代码

### Approach 1:

```java
class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        String tree1 = preorder(s);
        String tree2 = preorder(t);
        return tree1.indexOf(tree2) >= 0;
    }

    private String preorder(TreeNode t) {
        if (t == null) return "null";
        return "#" + t.val + " " + preorder(t.left) + " " + preorder(t.right);
    }
}
```

### Approach #2: By Comparison of Nodes

1、递归法自底向上Traverse

2、每层Traverse中，euqals当前节点值以及左右子节点是否相同

3、每层Traverse都会返回一个equals

**Complexity Analysis**

* Time complexity :&#x4F;*(\_m*∗*n). In worst case(skewed tree) `traverse` function takes \_O*(*m*∗*n*) time.
* Space complexity : &#x4F;*(\_n*). The depth of the recursion tree can go upto &#x6E;*. \_n* refers to the number of nodes in *s*.

```java
class Solution {
  public boolean isSubtree(TreeNode s, TreeNode t) {
    return traverse(s, t);
  }

  public boolean traverse(TreeNode s, TreeNode t) {
    return s != null && (equals(s, t) || traverse(s.left, t) || traverse(s.right, t));
  }

  public boolean equals(TreeNode x, TreeNode y) {
    if (x == null && y == null) return true;
    if (x == null || y == null) return false;

    return x.val == y.val 
      && equals(x.left, y.left) 
      && equals(x.right, y.right);
  }

}
```