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题目: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)
思路1:直接遍历两棵树返回序列,判断序列是否是子序列
思路2:比较AB根节点,如果不同,再递归比较A的左子树右子树和B
/**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(root1==null || root2==null) return false; return visit(root1).contains(visit(root2)); } public String visit(TreeNode root){ if(root==null)return ""; StringBuffer sb=new StringBuffer(); sb.append(root.val); sb.append(visit(root.left)); sb.append(visit(root.right)); return sb.toString(); }}
思路2
public class Solution { public boolean HasSubtree(TreeNode root1,TreeNode root2) { if (root1 == null || root2 == null) return false; return isSubTree(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2); } public boolean isSubTree(TreeNode pRootA, TreeNode pRootB){ if (pRootB == null) return true; if (pRootA == null) return false; if (pRootB.val == pRootA.val) { return isSubTree(pRootA.left, pRootB.left) && isSubTree(pRootA.right, pRootB.right); } else return false; }}
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