572. Subtree of Another Tree (Easy)

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.

Solution 1: Recursive, preorder O(m * n); O(m + n)

我们先从s的根结点开始，跟t比较，如果两棵树完全相同，那么返回true，否则就分别对s的左子结点和右子结点调用递归再次来判断是否相同，只要有一个返回true了，就表示可以找得到。

TC : O(m∗n). In worst case(skewed tree) traverse function takes O(m∗n) time. 就算不是倾斜树，是满二叉树，复杂度也是O(m * n)，因为最坏情况下，要遍历完s的所有节点，每遍历一个s节点，需要遍历完t的所有节点。

    public boolean isSubtree(TreeNode s, TreeNode t) {
        if (s == null) {
            return false;
        }
        if (isSame(s, t)) {
            return true;
        }

        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }

    private boolean isSame(TreeNode s, TreeNode t) { //和第100题代码一样
        if (s == null || t == null) {
            return s == t;
        }
        if (s.val != t.val) {
            return false;
        }
        return isSame(s.left, t.left) && isSame(s.right, t.right);
    }