Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its minimum depth = 2.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua, 4 ms
class Solution {
public:
  int minDepth(TreeNode* root) {
    if (!root) return 0;
    if (!root->left && !root->right) return 1;
    int l = root->left ? minDepth(root->left) : INT_MAX;
    int r = root->right ? minDepth(root->right) : INT_MAX;
    return min(l, r) + 1;
  }
};

class Solution:
  def minDepth(self, root):
    if not root: return 0
    if not root.left and not root.right: return 1
    l = self.minDepth(root.left)
    r = self.minDepth(root.right)
    if not root.left: return 1 + r
    if not root.right: return 1 + l
    return 1 + min(l, r)

Related Problem

• 花花酱 LeetCode 104. Maximum Depth of Binary Tree

The post 花花酱 LeetCode 111. Minimum Depth of Binary Tree appeared first on Huahua's Tech Road.

Given a n-ary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

For example, given a 3-ary tree:

We should return its max depth, which is 3.

Note:

1. The depth of the tree is at most 1000.
2. The total number of nodes is at most 5000.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua
// Running time: 40 ms
class Solution {
public:
  int maxDepth(Node* root) {
    if (root == nullptr) return 0;
    int depth = 0;
    for (auto child : root->children)
      depth = max(depth, maxDepth(child));
    return depth + 1;
  }
};

Related Problems

• 花花酱 LeetCode 590. N-ary Tree Postorder Traversal
• 花花酱 LeetCode 589. N-ary Tree Preorder Traversal

The post 花花酱 LeetCode 559. Maximum Depth of N-ary Tree appeared first on Huahua's Tech Road.

Problem

Given a binary tree rooted at root, the depth of each node is the shortest distance to the root.

A node is deepest if it has the largest depth possible among any node in the entire tree.

The subtree of a node is that node, plus the set of all descendants of that node.

Return the node with the largest depth such that it contains all the deepest nodes in it’s subtree.

Example 1:

Input: [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation:



We return the node with value 2, colored in yellow in the diagram.
The nodes colored in blue are the deepest nodes of the tree.
The input "[3, 5, 1, 6, 2, 0, 8, null, null, 7, 4]" is a serialization of the given tree.
The output "[2, 7, 4]" is a serialization of the subtree rooted at the node with value 2.
Both the input and output have TreeNode type.

Note:

• The number of nodes in the tree will be between 1 and 500.
• The values of each node are unique.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 0 ms
class Solution {
public:
  TreeNode* subtreeWithAllDeepest(TreeNode* root) {
    return depth(root).second;
  }
private:
  pair<int, TreeNode*> depth(TreeNode* root) {
    if (root == nullptr)
      return {-1, nullptr};
    auto l = depth(root->left);
    auto r = depth(root->right);
    int dl = l.first;
    int dr = r.first;
    return { max(dl, dr) + 1, 
             dl == dr ? root : (dl > dr) ? l.second : r.second};
  }
};

v2

// Author: Huahua
// Running time: 0 ms
class Solution {
public:
  TreeNode* subtreeWithAllDeepest(TreeNode* root) {
    TreeNode* ans;
    depth(root, &ans);
    return ans;
  }
private:
  int depth(TreeNode* root, TreeNode** s) {
    if (root == nullptr)
      return -1;
    TreeNode* sl;
    TreeNode* sr;
    int l = depth(root->left, &sl);
    int r = depth(root->right, &sr);
    *s = (l == r) ? root : ((l > r) ? sl : sr);
    return max(l, r) + 1;
  }
};

Python3

"""
Author: Huahua
Running time: 40 ms
"""
class Solution:
  def subtreeWithAllDeepest(self, root):
    def solve(root):
      if not root: return (-1, None)
      dl, sl = solve(root.left)
      dr, sr = solve(root.right)
      if dl == dr: return (dl + 1, root)
      return (dl + 1, sl) if dl > dr else (dr + 1, sr)
    return solve(root)[1]

The post 花花酱 LeetCode 865. Smallest Subtree with all the Deepest Nodes appeared first on Huahua's Tech Road.