logseq_notes/pages/OJ notes/pages/Leetcode Maximum-Depth-Of-Binary-Tree.md

Leetcode Maximum-Depth-Of-Binary-Tree

2022-07-05 09:25

Algorithms:

#algorithm #BFS

Data structures:

#DS #binary_tree

Difficulty:

#coding_problems #difficulty_easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

  ---

Problem

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Examples

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: 3

Example 2:

Input: root = [1,null,2] Output: 2

Constraints

• The number of nodes in the tree is in the range [0, 104].
• -100 <= Node.val <= 100

Thoughts

[!summary] This problem can be solved by #BFS or #DFS

BFS way: Simply log the level in each while iteration.

DFS way: (Popular) Use recursion:

• Base Case:
  • root == nullptr: return 0;
• maxDepth(root) = max(maxDepth(root->left), maxDepth(root->right))

Solution

DFS Recursion:

/**
 * 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 maxDepth(TreeNode *root) {
  // DFS
  if (!root) {
    return 0;
  }

  return max(maxDepth(root->left), maxDepth(root->right)) + 1;
}
};

BFS:

/**
 * 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 maxDepth(TreeNode *root) {
  // BFS
  int levels = 0;
  queue<TreeNode *> pending;
  TreeNode *ptr = root;
  if (ptr)
    pending.push(ptr);

  while (!pending.empty()) {
    levels++;
    for (int i = 0, size = pending.size(); i < size; i++) {
      ptr = pending.front();
      pending.pop();

      if (ptr->left)
        pending.push(ptr->left);
      if (ptr->right)
        pending.push(ptr->right);
    }
  }

  return levels;
}
};