A binary tree is univalued if every node in the tree has the same value.
Return true if and only if the given tree is univalued.
Example 1:
Input: [1,1,1,1,1,null,1] Output: true
Example 2:
Input: [2,2,2,5,2] Output: false
Note:
- The number of nodes in the given tree will be in the range
[1, 100]. - Each node's value will be an integer in the range
[0, 99].
题目描述：大概意思就是问我们给定一棵树，判断这棵树上的所有节点的值是不是相同的，相同即为 true ，不相同为 false 。
题目分析：判断一棵树的所有节点的值是不是相同的，可以分为以下几个条件：
- 节点是否为空
- 左子节点和父节点是否相同
- 右子节点和父节点是否相同
- 左子节点和右子节点是否相同
根据这个思路我们可以解决这个问题。
python 代码：
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def isUnivalTree(self, root): """ :type root: TreeNode :rtype: bool """ left_correct = not root.left or root.val == root.left.val and self.isUnivalTree(root.left) right_correct = not root.right or root.val == root.right.val and self.isUnivalTree(root.right) return left_correct and right_correct
C++ 代码：
/** * 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 temp; bool flag = true; bool isUnivalTree(TreeNode* root) { if(!root){ return true; } temp = root->val; travelTree(root); return flag; } void travelTree(TreeNode* root){ if(root){ travelTree(root->left); travelTree(root->right); if(flag){ flag = root->val == temp ? true : false; } } } };
