LeetCode 965. Univalued Binary Tree

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:

1. The number of nodes in the given tree will be in the range [1, 100].
2. 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;
            }
        }
    }

};