1339:分裂二叉树的最大乘积
This commit is contained in:
轩辕龙儿
parent
f91664d203
commit
7feb720997
@ -0,0 +1,120 @@
|
||||
//<p>给你一棵二叉树，它的根为 <code>root</code> 。请你删除 1 条边，使二叉树分裂成两棵子树，且它们子树和的乘积尽可能大。</p>
|
||||
//
|
||||
//<p>由于答案可能会很大，请你将结果对 10^9 + 7 取模后再返回。</p>
|
||||
//
|
||||
//<p> </p>
|
||||
//
|
||||
//<p><strong>示例 1：</strong></p>
|
||||
//
|
||||
//<p><strong><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/02/sample_1_1699.png" style="height: 200px; width: 495px;"></strong></p>
|
||||
//
|
||||
//<pre><strong>输入：</strong>root = [1,2,3,4,5,6]
|
||||
//<strong>输出：</strong>110
|
||||
//<strong>解释：</strong>删除红色的边，得到 2 棵子树，和分别为 11 和 10 。它们的乘积是 110 （11*10）
|
||||
//</pre>
|
||||
//
|
||||
//<p><strong>示例 2：</strong></p>
|
||||
//
|
||||
//<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/02/sample_2_1699.png" style="height: 200px; width: 495px;"></p>
|
||||
//
|
||||
//<pre><strong>输入：</strong>root = [1,null,2,3,4,null,null,5,6]
|
||||
//<strong>输出：</strong>90
|
||||
//<strong>解释：</strong>移除红色的边，得到 2 棵子树，和分别是 15 和 6 。它们的乘积为 90 （15*6）
|
||||
//</pre>
|
||||
//
|
||||
//<p><strong>示例 3：</strong></p>
|
||||
//
|
||||
//<pre><strong>输入：</strong>root = [2,3,9,10,7,8,6,5,4,11,1]
|
||||
//<strong>输出：</strong>1025
|
||||
//</pre>
|
||||
//
|
||||
//<p><strong>示例 4：</strong></p>
|
||||
//
|
||||
//<pre><strong>输入：</strong>root = [1,1]
|
||||
//<strong>输出：</strong>1
|
||||
//</pre>
|
||||
//
|
||||
//<p> </p>
|
||||
//
|
||||
//<p><strong>提示：</strong></p>
|
||||
//
|
||||
//<ul>
|
||||
// <li>每棵树最多有 <code>50000</code> 个节点，且至少有 <code>2</code> 个节点。</li>
|
||||
// <li>每个节点的值在 <code>[1, 10000]</code> 之间。</li>
|
||||
//</ul>
|
||||
//<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>二叉树</li></div></div><br><div><li>👍 72</li><li>👎 0</li></div>
|
||||
package leetcode.editor.cn;
|
||||
|
||||
import com.code.leet.entiy.TreeNode;
|
||||
|
||||
import java.util.HashMap;
|
||||
import java.util.Map;
|
||||
|
||||
// 1339:分裂二叉树的最大乘积
|
||||
public class MaximumProductOfSplittedBinaryTree {
|
||||
public static void main(String[] args) {
|
||||
Solution solution = new MaximumProductOfSplittedBinaryTree().new Solution();
|
||||
// TO TEST
|
||||
}
|
||||
//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 {
|
||||
public int maxProduct(TreeNode root) {
|
||||
map = new HashMap<>();
|
||||
getSums(root);
|
||||
dfs(root, map.get(root));
|
||||
return (int) (max % 1000000007);
|
||||
}
|
||||
|
||||
Map<TreeNode, Integer> map;
|
||||
long max = 0;
|
||||
|
||||
private int getSums(TreeNode node) {
|
||||
if (node == null) {
|
||||
return 0;
|
||||
}
|
||||
int sum = node.val;
|
||||
if (node.left != null) {
|
||||
sum += getSums(node.left);
|
||||
}
|
||||
if (node.right != null) {
|
||||
sum += getSums(node.right);
|
||||
}
|
||||
map.put(node, sum);
|
||||
return sum;
|
||||
}
|
||||
|
||||
private void dfs(TreeNode node, int sum) {
|
||||
if (node == null) {
|
||||
return;
|
||||
}
|
||||
if (node.left != null) {
|
||||
int num1 = map.get(node.left);
|
||||
max = Math.max(max, (long) num1 * (sum - num1));
|
||||
dfs(node.left, sum);
|
||||
}
|
||||
if (node.right != null) {
|
||||
int num1 = map.get(node.right);
|
||||
max = Math.max(max, (long) num1 * (sum - num1));
|
||||
dfs(node.right, sum);
|
||||
}
|
||||
}
|
||||
}
|
||||
//leetcode submit region end(Prohibit modification and deletion)
|
||||
|
||||
}
|
||||
@ -0,0 +1,45 @@
|
||||
<p>给你一棵二叉树，它的根为 <code>root</code> 。请你删除 1 条边，使二叉树分裂成两棵子树，且它们子树和的乘积尽可能大。</p>
|
||||
|
||||
<p>由于答案可能会很大，请你将结果对 10^9 + 7 取模后再返回。</p>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>示例 1：</strong></p>
|
||||
|
||||
<p><strong><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/02/sample_1_1699.png" style="height: 200px; width: 495px;"></strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>root = [1,2,3,4,5,6]
|
||||
<strong>输出：</strong>110
|
||||
<strong>解释：</strong>删除红色的边，得到 2 棵子树，和分别为 11 和 10 。它们的乘积是 110 （11*10）
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 2：</strong></p>
|
||||
|
||||
<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/02/sample_2_1699.png" style="height: 200px; width: 495px;"></p>
|
||||
|
||||
<pre><strong>输入：</strong>root = [1,null,2,3,4,null,null,5,6]
|
||||
<strong>输出：</strong>90
|
||||
<strong>解释：</strong>移除红色的边，得到 2 棵子树，和分别是 15 和 6 。它们的乘积为 90 （15*6）
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 3：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>root = [2,3,9,10,7,8,6,5,4,11,1]
|
||||
<strong>输出：</strong>1025
|
||||
</pre>
|
||||
|
||||
<p><strong>示例 4：</strong></p>
|
||||
|
||||
<pre><strong>输入：</strong>root = [1,1]
|
||||
<strong>输出：</strong>1
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
|
||||
<p><strong>提示：</strong></p>
|
||||
|
||||
<ul>
|
||||
<li>每棵树最多有 <code>50000</code> 个节点，且至少有 <code>2</code> 个节点。</li>
|
||||
<li>每个节点的值在 <code>[1, 10000]</code> 之间。</li>
|
||||
</ul>
|
||||
<div><div>Related Topics</div><div><li>树</li><li>深度优先搜索</li><li>二叉树</li></div></div><br><div><li>👍 72</li><li>👎 0</li></div>
|
||||
Loading…
Reference in New Issue
Block a user