Question

250. Count Univalue Subtrees

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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

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

/**
 * 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;
};