剑指offer 26.树的子结构。简单易懂0ms

class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if(B == null || A == null) return false;
        if(A.val == B.val && dfs(A.left, B.left) && dfs(A.right, B.right)) return true;
        // 判断B在左子树还是右子树
        return isSubStructure(A.left, B) || isSubStructure(A.right, B);
    }

    public boolean dfs(TreeNode node1, TreeNode node2){
        // 先对node2进行判断，为空说明是子结构
        // 然后对node1进行判断，如果为空说明不是
        if(node2 == null) return true;
        if(node1 == null) return false;
        if(node1.val == node2.val){
            return dfs(node1.left, node2.left)&&dfs(node1.right, node2.right);
        }else{
            return false;
        }
    }
}