一. 思路

若树 B 是树 A 的子结构，则子结构的根节点可能为树 A 的任意一个节点。详情看leetcode题解：面试题26. 树的子结构（先序遍历 + 包含判断，清晰图解）

二. 代码

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
class Solution {
    public boolean HasSubtree(TreeNode A, TreeNode B) {
        return (A != null && B != null) && (recur(A, B) || HasSubtree(A.left, B) || HasSubtree(A.right, B));
    }
    boolean recur(TreeNode A, TreeNode B) {
        if(B == null) return true;
        if(A == null || A.val != B.val) return false;
        return recur(A.left, B.left) && recur(A.right, B.right);
    }
}