问题：

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.

解决：

① 给定两个树s和t，判断t是否是s的一个子树。这里s的子树是指，一个树由s中的一个节点及其所有子节点组成，s也可以是自己本身的一个子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution { //28ms
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if(t == null) return true;
        if(s == null) return false;
        if(isSame(s,t)) return true;
        return isSubtree(s.left,t) || isSubtree(s.right,t);
    }
    public boolean isSame(TreeNode s,TreeNode t){
        if(s == null && t == null) return true;
        if(s == null || t == null) return false;
        if(s.val != t.val) return false;
        return isSame(s.left,t.left) && isSame(s.right,t.right);
    }
}

② 在discuss中看到以下解法

public class Solution { // 16ms
    public boolean isSubtree(TreeNode s, TreeNode t) {
        return isSubtree(s, t, true);
    }
    private boolean isSubtree(TreeNode s, TreeNode t, boolean isHead) {
        if (s == null && t == null) return true;
        if (s == null || t == null || (s.val != t.val && !isHead)) return false; //isHead表示是否以当前节点为根节点比较子树的
        if (s.val == t.val) {
            boolean left = isSubtree(s.left, t.left, false);
            boolean right = isSubtree(s.right, t.right, false);
            if (left && right) {
                return true;
            } else {
                left = isSubtree(s.left, t, true);
                right = isSubtree(s.right, t, true);
                return left || right;
            }
        } else {
            boolean left = isSubtree(s.left, t, true);
            boolean right = isSubtree(s.right, t, true);
            return left || right;
        }
    }
}