[LeetCode] 250. Count Univalue Subtrees 计算唯一值子树的个数

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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);
}
};