Sub-tree - cattle off network

 

Two binary inputs A, B, B is judged not substructure A's. (Ps: we agreed empty tree is not a tree any substructure)

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

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

    }

}
*/
public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        boolean  result = false;
        if(root1 == null || root2 == null) return false;
        else{
            if(root1.val == root2.val) result = doequal(root1,root2);
            //判断上一个if是否得出true
            if(!result) result = HasSubtree(root1.left,root2);//遍历树，在1中找到与2的跟相等的元素。
            if(!result) result = HasSubtree(root1.right,root2);
        }
        return result;
    }
    public boolean doequal(TreeNode root1,TreeNode root2) {
        //拟子目录先遍历完，即为真
        if(root2 == null) return true;
        //拟父目录先走完，即为假
        if(root1 == null) return false;
        //对比过程中元素不相等，为假
        if(root1.val != root2.val) return false;
        //把拟子目录与父目录上元素对比
        return (doequal(root1.right,root2.right)&&doequal(root1.left,root2.left));
    }
}