标签：树的子结构 container val aci 开始 help solution 提前 情况下
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）
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helper(root1.left,root2.left)&&helper(root1.right,root2.right) |
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helper(root1.left,root2)||helper(root1.right,root2) |
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helper(root1.left,root2)||helper(root1.right,root2) |
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/**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) { if(root2==null) return false; return helper(root1,root2); } public boolean helper(TreeNode root1,TreeNode root2) { if(root2==null) return true; if(root1==null) return false; if(root1.val==root2.val) return (helper(root1.left,root2.left)&&helper(root1.right,root2.right))||(helper(root1.left,root2)||helper(root1.right,root2)); else return helper(root1.left,root2)||helper(root1.right,root2); }} |
标签：树的子结构 container val aci 开始 help solution 提前 情况下
原文地址：https://www.cnblogs.com/cold-windy/p/11534398.html
