A unival tree (which stands for "universal value") is a tree where all nodes under it have the same value.
Given the root to a binary tree, count the number of unival subtrees.
For example, the following tree has 5 unival subtrees:
0
/ \
1 0
/ \
1 0
/ \
1 1
function Node(val) { return { val, left: null, right: null }; } const root = Node(0); root.left = Node(1); root.right = Node(0); root.right.left = Node(1); root.right.right = Node(0); root.right.left.left = Node(1); root.right.left.right = Node(1); function count_unival(root) { function helper(root) { let total_count = 0; let is_unival = true; // Base case 1: if current node is null, then return if (root == null) { return [0, true]; } // Base case 2: if current node is not null, but its children node // are null, then count this node as usb-unvial tree if (root.left === null && root.right === null) { return [1, true]; } // Base case 1 & Base case 2 can keep just one, it should still works // Do the Recursion let [left_count, is_left_unival] = helper(root.left); let [right_count, is_right_unival] = helper(root.right); // we need to consider whether the whole tree // root + left tree + right tree are unvial // the way to do it just compare root with its left and right node // whether they are the same if both left tree and right tree are // unival tree. if (!is_left_unival || !is_right_unival) { is_unival = false; } if (root.left !== null && root.val !== root.left.val) { is_unival = false; } if (root.right !== null && root.val !== root.right.val) { is_unival = false; } // If the whole tree are unival tree, then the final result // should + 1 if (is_unival) { return [left_count + right_count + 1, is_unival]; } else { return [left_count + right_count, is_unival]; } } const [total_count, is_unival] = helper(root); return [total_count, is_unival]; } const res = count_unival(root); console.log( `Whole tree is${ res[1] ? "" : "n't" } unival tree, total counts for sub-unival tree is ${res[0]}` ); // Whole tree isn't unival tree, total count is 5