题解 | #树的子结构#

public class Solution {
    boolean[] res = new boolean[] {false};
    public boolean HasSubtree(TreeNode root1, TreeNode root2) {
        if(root2==null) return false;
        if (root1 == null) return false;
    
        if (compareFromRootNode(root1, root2)) {
            res[0] = true;
        }

        HasSubtree(root1.left, root2);
        HasSubtree(root1.right, root2);
        if (res[0]) {
            return true;
        } else {
            return false;
        }
    }

    private boolean compareFromRootNode(TreeNode root1, TreeNode root2) {
      
        if (root2 == null) return true;
        if (root1==null) return false;
        if (root1.val != root2.val) return false;
        return compareFromRootNode(root1.left, root2.left) &&
               compareFromRootNode(root1.right, root2.right);
    }


    
}

缝缝补补的屎山代码