标签:for http amp bsp 相同 tco lag bool 判断
https://leetcode-cn.com/problems/subtree-of-another-tree/
我们需要借助一个helper方法,这个思想就是双递归,即判断每个大树的节点和小树的根节点的值是否相等,如果相同就进入子树判断方法。否则就在大树节点的左节点和右节点查找。
helper方法中,如果小树为空大树不为空或者两个值不相等,就返回false,如果两个都为空才返回true,否则就继续递归寻找左子树和右子树。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public boolean isSubtree(TreeNode s, TreeNode t) { if(s == null){ return false; } boolean flag = false; if(s.val == t.val){ flag = helper(s.left,t.left) && helper(s.right, t.right); } return flag?flag:isSubtree(s.left,t) || isSubtree(s.right,t); } private boolean helper (TreeNode s, TreeNode t){ if(t == null){ if(s != null){ return false; } return true; } if(s == null || s.val != t.val){ return false; } return helper(s.left,t.left)&&helper(s.right,t.right); } }
标签:for http amp bsp 相同 tco lag bool 判断
原文地址:https://www.cnblogs.com/ZJPaang/p/12842199.html