树的子结构 Posted on 2019-09-19 | In 剑指offer | | reads times 树的子结构题目描述 输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构) 重复两步,采用递归 根节点是否相同 相同检查剩余节点 123456789101112131415161718function HasSubtree(pRoot1, pRoot2){ // write code here let res=false; if(pRoot1===null)return false; if(pRoot2===null)return false; if(pRoot1.val==pRoot2.val) res=tree1hastree2(pRoot1,pRoot2) if(!res)res=HasSubtree(pRoot1.left,pRoot2) if(!res)res=HasSubtree(pRoot1.right,pRoot2) return res;}function tree1hastree2(pRoot1,pRoot2){ if(pRoot2===null)return true; if(pRoot1===null)return false; if(pRoot1.val!==pRoot2.val)return false; return tree1hastree2(pRoot1.left,pRoot2.left)&&tree1hastree2(pRoot1.right,pRoot2.right);} Post author: GoldMiner Xun Post link: https://goldminerxun.github.io/2019/09/19/%E5%89%91%E6%8C%87offer%20JavaScript%E7%89%88%20(17)/ Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 3.0 unless stating additionally.