652:寻找重复的子树

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huangge1199 2021-09-03 10:54:17 +08:00
parent 8680fe9701
commit a2e1eed847
2 changed files with 119 additions and 0 deletions

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//给定一棵二叉树返回所有重复的子树对于同一类的重复子树你只需要返回其中任意一棵的根结点即可
//
// 两棵树重复是指它们具有相同的结构以及相同的结点值
//
// 示例 1
//
// 1
// / \
// 2 3
// / / \
// 4 2 4
// /
// 4
//
//
// 下面是两个重复的子树
//
// 2
// /
// 4
//
//
//
//
// 4
//
//
// 因此你需要以列表的形式返回上述重复子树的根结点
// Related Topics 深度优先搜索 广度优先搜索 二叉树 👍 303 👎 0
package leetcode.editor.cn;
import com.code.leet.entiy.TreeNode;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
//652:寻找重复的子树
class FindDuplicateSubtrees {
public static void main(String[] args) {
//测试代码
Solution solution = new FindDuplicateSubtrees().new Solution();
}
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
/**
* 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 {
Map<String, Integer> count;
List<TreeNode> ans;
public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
count = new HashMap();
ans = new ArrayList();
collect(root);
return ans;
}
public String collect(TreeNode node) {
if (node == null) {
return "#";
}
String serial = node.val + "," + collect(node.left) + "," + collect(node.right);
count.put(serial, count.getOrDefault(serial, 0) + 1);
if (count.get(serial) == 2) {
ans.add(node);
}
return serial;
}
}
//leetcode submit region end(Prohibit modification and deletion)
}

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<p>给定一棵二叉树,返回所有重复的子树。对于同一类的重复子树,你只需要返回其中任意<strong>一棵</strong>的根结点即可。</p>
<p>两棵树重复是指它们具有相同的结构以及相同的结点值。</p>
<p><strong>示例 1</strong></p>
<pre> 1
/ \
2 3
/ / \
4 2 4
/
4
</pre>
<p>下面是两个重复的子树:</p>
<pre> 2
/
4
</pre>
<p></p>
<pre> 4
</pre>
<p>因此,你需要以列表的形式返回上述重复子树的根结点。</p>
<div><div>Related Topics</div><div><li></li><li>深度优先搜索</li><li>广度优先搜索</li><li>二叉树</li></div></div><br><div><li>👍 303</li><li>👎 0</li></div>