965. 单值二叉树

题目描述

如果二叉树每个节点都具有相同的值，那么该二叉树就是单值二叉树。

只有给定的树是单值二叉树时，才返回 true；否则返回 false。

示例 1：

输入：[1,1,1,1,1,null,1]
输出：true

示例 2：

输入：[2,2,2,5,2]
输出：false

提示：

给定树的节点数范围是 [1, 100]。
每个节点的值都是整数，范围为 [0, 99] 。

解法

方法一

Python3JavaC++GoTypeScriptRust

# 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 isUnivalTree(self, root: TreeNode) -> bool:
        def dfs(node):
            if node is None:
                return True
            return node.val == root.val and dfs(node.left) and dfs(node.right)

        return dfs(root)

/**
 * 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 {
    public boolean isUnivalTree(TreeNode root) {
        return dfs(root, root.val);
    }

    private boolean dfs(TreeNode root, int val) {
        if (root == null) {
            return true;
        }
        return root.val == val && dfs(root.left, val) && dfs(root.right, val);
    }
}

/**
 * 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:
    bool isUnivalTree(TreeNode* root) {
        return dfs(root, root->val);
    }

    bool dfs(TreeNode* root, int val) {
        if (!root) return true;
        return root->val == val && dfs(root->left, val) && dfs(root->right, val);
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isUnivalTree(root *TreeNode) bool {
    var dfs func(*TreeNode) bool
    dfs = func(node *TreeNode) bool {
        if node == nil {
            return true
        }
        return node.Val == root.Val && dfs(node.Left) && dfs(node.Right)
    }
    return dfs(root)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isUnivalTree(root: TreeNode | null): boolean {
    const val = root.val;
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return true;
        }
        return root.val === val && dfs(root.left) && dfs(root.right);
    };
    return dfs(root.left) && dfs(root.right);
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(val: i32, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
        if root.is_none() {
            return true;
        }
        let root = root.as_ref().unwrap().borrow();
        root.val == val && Self::dfs(val, &root.left) && Self::dfs(val, &root.right)
    }
    pub fn is_unival_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        let root = root.as_ref().unwrap().borrow();
        Self::dfs(root.val, &root.left) && Self::dfs(root.val, &root.right)
    }
}

965. 单值二叉树

题目描述

解法

方法一

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