250 Count Univalue Subtrees 統計同值子樹
Description:
Given the root of a binary tree, return the number of uni-value subtrees.
A uni-value subtree means all nodes of the subtree have the same value.
Example:
Example 1:
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Input: root = [5,1,5,5,5,null,5]
Output: 4
Example 2:
Input: root = []
Output: 0
Example 3:
Input: root = [5,5,5,5,5,null,5]
Output: 6
Constraints:
The number of the node in the tree will be in the range [0, 1000].
-1000 <= Node.val <= 1000
題目描述:
給定一個二叉樹,統計該二叉樹數值相同的子樹個數。
同值子樹是指該子樹的所有節點都擁有相同的數值。
示例:
輸入: root = [5,1,5,5,5,null,5]
5
/ \
1 5
/ \ \
5 5 5
輸出: 4
思路:
遞歸後序遍歷
當遍歷到空結點時返回 true
先遍歷左子樹, 再遍歷右子樹, 最後遍歷根結點
如果爲葉子結點爲 true, 計數器加 1
從葉子結點往上遍歷, 子樹值完全相同的計數器加 1
即左子樹右子樹和根結點值相等
時間複雜度爲 O(n), 空間複雜度爲 O(n), 最差情況樹退化爲鏈表
代碼:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
int countUnivalSubtrees(TreeNode* root)
{
result = 0;
dfs(root);
return result;
}
private:
int result;
bool dfs(TreeNode* root)
{
if (!root) return true;
bool left = dfs(root -> left), right = dfs(root -> right), cur = (!root -> left or root -> val == root -> left -> val) and (!root -> right or root -> val == root -> right -> val);
if (cur and left and right) ++result;
return cur and left and right;
}
};
Java:
/**
* 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 {
private int result = 0;
public int countUnivalSubtrees(TreeNode root) {
dfs(root);
return result;
}
private boolean dfs(TreeNode root) {
if (root == null) return true;
boolean left = dfs(root.left), right = dfs(root.right), cur = (root.left == null || root.val == root.left.val) && (root.right == null || root.val == root.right.val);
if (cur && left && right) ++result;
return cur && left && right;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countUnivalSubtrees(self, root: Optional[TreeNode]) -> int:
count = 0
def dfs(root: Optional[TreeNode]) -> bool:
nonlocal count
if not root:
return True
left, right = dfs(root.left), dfs(root.right)
count += (cur := (not root.left or root.val == root.left.val) and (not root.right or root.val == root.right.val)) and left and right
return cur and left and right
dfs(root)
return count