LeetCode 题解(265) : Count Univalue Subtrees

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题目：

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.

题解：

递归。

Java版：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public int countUnivalSubtrees(TreeNode root) {unival(root);return count;}private boolean unival(TreeNode root) {if(root == null)return true;if(root.left ==null && root.right == null) {count++;return true;}boolean left = unival(root.left);boolean right = unival(root.right);if(left && right && (root.left == null || root.left.val == root.val) && (root.right == null || root.right.val == root.val)) {count++;return true;}return false;}private int count = 0;
}

Python版：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def __init__(self):self.count = 0def countUnivalSubtrees(self, root):""":type root: TreeNode:rtype: int"""self.unival(root)return self.countdef unival(self, root):if root == None:return True if root.left == None and root.right == None:self.count += 1return Trueleft = self.unival(root.left)right = self.unival(root.right)if left and right and (root.left == None or root.left.val == root.val) and (root.right == None or root.right.val == root.val):self.count += 1return Trueelse:return False

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