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250. Count Univalue Subtrees πŸ”’

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 1:

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

Solutions

Solution 1

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# 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:
        def dfs(root):
            if root is None:
                return True
            l, r = dfs(root.left), dfs(root.right)
            if not l or not r:
                return False
            a = root.val if root.left is None else root.left.val
            b = root.val if root.right is None else root.right.val
            if a == b == root.val:
                nonlocal ans
                ans += 1
                return True
            return False

        ans = 0
        dfs(root)
        return ans

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/**
 * 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 ans;

    public int countUnivalSubtrees(TreeNode root) {
        dfs(root);
        return ans;
    }

    private boolean dfs(TreeNode root) {
        if (root == null) {
            return true;
        }
        boolean l = dfs(root.left);
        boolean r = dfs(root.right);
        if (!l || !r) {
            return false;
        }
        int a = root.left == null ? root.val : root.left.val;
        int b = root.right == null ? root.val : root.right.val;
        if (a == b && b == root.val) {
            ++ans;
            return true;
        }
        return false;
    }
}

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/**
 * 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) {
        int ans = 0;
        function<bool(TreeNode*)> dfs = [&](TreeNode* root) -> bool {
            if (!root) {
                return true;
            }
            bool l = dfs(root->left);
            bool r = dfs(root->right);
            if (!l || !r) {
                return false;
            }
            int a = root->left ? root->left->val : root->val;
            int b = root->right ? root->right->val : root->val;
            if (a == b && b == root->val) {
                ++ans;
                return true;
            }
            return false;
        };
        dfs(root);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func countUnivalSubtrees(root *TreeNode) (ans int) {
    var dfs func(*TreeNode) bool
    dfs = func(root *TreeNode) bool {
        if root == nil {
            return true
        }
        l, r := dfs(root.Left), dfs(root.Right)
        if !l || !r {
            return false
        }
        if root.Left != nil && root.Left.Val != root.Val {
            return false
        }
        if root.Right != nil && root.Right.Val != root.Val {
            return false
        }
        ans++
        return true
    }
    dfs(root)
    return
}

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/**
 * 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 countUnivalSubtrees(root: TreeNode | null): number {
    let ans: number = 0;
    const dfs = (root: TreeNode | null): boolean => {
        if (root == null) {
            return true;
        }
        const l: boolean = dfs(root.left);
        const r: boolean = dfs(root.right);
        if (!l || !r) {
            return false;
        }
        if (root.left != null && root.left.val != root.val) {
            return false;
        }
        if (root.right != null && root.right.val != root.val) {
            return false;
        }
        ++ans;
        return true;
    };
    dfs(root);
    return ans;
}

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/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var countUnivalSubtrees = function (root) {
    let ans = 0;
    const dfs = root => {
        if (!root) {
            return true;
        }
        const l = dfs(root.left);
        const r = dfs(root.right);
        if (!l || !r) {
            return false;
        }
        if (root.left && root.left.val !== root.val) {
            return false;
        }
        if (root.right && root.right.val !== root.val) {
            return false;
        }
        ++ans;
        return true;
    };
    dfs(root);
    return ans;
};

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