Given a binary tree, count the number of uni-value subtrees.
A Uni-value subtree means all nodes of the subtree have the same value.
For example:
Given binary tree,
5 / 1 5 / \ 5 5 5
return 4
.
给一个二叉树,求唯一值子树的个数。唯一值子树的所有节点具有相同值。
解法:递归
Java:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { int count = 0; public int countUnivalSubtrees(TreeNode root) { if (root == null) return 0; isUnival(root); return count; } private boolean isUnival(TreeNode root) { if (root == null) return true; if (isUnival(root.left) & isUnival(root.right)) { if (root.left != null && root.left.val != root.val) return false; if (root.right != null && root.right.val != root.val) return false; count++; return true; } return false; } }
Java:
public class Solution { public int countUnivalSubtrees(TreeNode root) { int[] count = new int[] {0}; isUnivalSubtrees(root,count); return count[0]; } private boolean isUnivalSubtrees(TreeNode root, int[] count) { if(root == null) return true; boolean left = isUnivalSubtrees(root.left, count); boolean right = isUnivalSubtrees(root.right, count); if(left && right) { if(root.left != null && root.left.val != root.val) { return false; } if(root.right != null && root.right.val != root.val) { return false; } count[0]++; return true; } return false; } }
Python:
# Time: O(n) # Space: O(h) class Solution(object): # @param {TreeNode} root # @return {integer} def countUnivalSubtrees(self, root): [is_uni, count] = self.isUnivalSubtrees(root, 0); return count; def isUnivalSubtrees(self, root, count): if not root: return [True, count] [left, count] = self.isUnivalSubtrees(root.left, count) [right, count] = self.isUnivalSubtrees(root.right, count) if self.isSame(root, root.left, left) and self.isSame(root, root.right, right): count += 1 return [True, count] return [False, count] def isSame(self, root, child, is_uni): return not child or (is_uni and root.val == child.val)
C++:
// Time: O(n) // Space: O(h) /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int countUnivalSubtrees(TreeNode* root) { int count = 0; isUnivalSubtrees(root, &count); return count; } bool isUnivalSubtrees(TreeNode* root, int *count) { if (root == nullptr) { return true; } bool left = isUnivalSubtrees(root->left, count); bool right = isUnivalSubtrees(root->right, count); if (isSame(root, root->left, left) && isSame(root, root->right, right)) { ++(*count); return true; } return false; } bool isSame(TreeNode* root, TreeNode* child, bool is_uni) { return child == nullptr || (is_uni && root->val == child->val); } };
类似题目:
[LeetCode] 687. Longest Univalue Path 最长唯一值路径
原文地址:https://www.cnblogs.com/lightwindy/p/9795764.html
时间: 2024-11-05 21:48:31