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# 572.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:

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 :O(_mn). In worst case(skewed tree) `traverse` function takes _O(mn) time.
• Space complexity : O(_n). The depth of the recursion tree can go upto n. _n refers to the number of nodes in s.
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);
}
}