[编程题]树的子结构

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

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

    }

}
*/
/*
递归运算
注意各个if条件语句的先后顺序
*/
public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        boolean result=false;
        //递归终止条件
        if(root1!=null&&root2!=null){

            //判断树1与树2的头结点相等吗
            if(root1.val==root2.val){
                result=subtreeIsEqual(root1,root2);
            }

            if(!result){
            //判断树1的左子树与树2的子树相等吗
                result=HasSubtree(root1.left,root2);
            }

            if(!result){
            //判断树1的右子树与树2的子树相等吗
                result=HasSubtree(root1.right,root2);
            }
        }
        return result;
        }

            public boolean subtreeIsEqual(TreeNode subroot1,TreeNode subroot2){
        //递归终止条件
            //先判断树2为空吗，因为subtreeIsEqual执行的条件是父节点的值分别相等
            if(subroot2==null){
            return true;
        }
        //再判断树1为空吗
        if(subroot1==null){
            return false;
        }

        //判断根节点
        if(subroot1.val!=subroot2.val){
            return false;
        }else{
            return subtreeIsEqual(subroot1.left,subroot2.left)&&subtreeIsEqual
                (subroot1.right,subroot2.right);
        }

    }

}