We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Tree: Height of a Binary Tree
  2. Discussions

Tree: Height of a Binary Tree

Sort by

recency

|

12 Discussions

|

  • perfetti_stefano
     + 0 comments

    How was this problem classified as requiring advanced problem solving? It is much easier than many supposedly easy problems in this preparation kit.

  • vadym_usatiuk
     + 0 comments

    Java 8, O(log n)

    public static int height(Node root) {
            if (root == null) {
                return -1;
            } else {
                return 1 + Math.max(height(root.left), height(root.right));
            }
        }

  • dariaabaklanova
     + 0 comments

    It seems it's broken on JS: it's comparing NaN with root.data preparing the Tree. and all the nodes are on the right then...

  • logan_r_smith0
     + 1 comment

    C++ using a stack and iteration instead of recursion.

        int height(Node* root) {
        struct Frame {
            Node* node{};
            int height = 0;
        };
        std::stack<Frame> stack;
        stack.push({root});
        int max = INT_MIN;
        while (!stack.empty()) {
            const auto top = stack.top();
            stack.pop();
            max = std::max(max, top.height);
            if (top.node->right)
                stack.push({top.node->right, top.height + 1});
            if (top.node->left)
                stack.push({top.node->left, top.height + 1});
        }
        return max;
    }

  • florenciopaucar
     + 0 comments
    def height(root):
    '''
    Solved with BFS, this is adapt to get shortest distances in tree with edges weight=1
    '''
    explored = {}
    explored[root.info] = 0
    q = deque()
    q.append(root)
    height_tree = 0
    while q:
        v = q.popleft()
        children = []
        if v.left:
            children.append(v.left)
        if v.right:
            children.append(v.right)
        for w in children:
            if not explored.get(w.info) and (w.info not in explored):
                explored[w.info] = explored[v.info] + 1
                height_tree = max(explored[w.info], height_tree)
                q.append(w)
    return height_tree