145 lines
2.6 KiB
Markdown
145 lines
2.6 KiB
Markdown
# Leetcode Maximum-Depth-Of-Binary-Tree
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#### 2022-07-05 09:25
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> ##### Algorithms:
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>
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> #algorithm #BFS
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>
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> ##### Data structures:
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>
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> #DS #binary_tree
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty-easy
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/maximum-depth-of-binary-tree/)
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---
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### Problem
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Given the `root` of a binary tree, return _its maximum depth_.
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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.
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#### Examples
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**Example 1:**
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**Input:** root = [3,9,20,null,null,15,7]
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**Output:** 3
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**Example 2:**
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**Input:** root = [1,null,2]
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**Output:** 2
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#### Constraints
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- The number of nodes in the tree is in the range `[0, 104]`.
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- `-100 <= Node.val <= 100`
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### Thoughts
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> [!summary]
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> This problem can be solved by #BFS or #DFS
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BFS way:
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Simply log the level in each while iteration.
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DFS way: (Popular)
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Use recursion:
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- Base Case:
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- root == nullptr: return 0;
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- maxDepth(root) = max(maxDepth(root->left), maxDepth(root->right))
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### Solution
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DFS Recursion:
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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* right(right) {}
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* };
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*/
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class Solution {
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public:
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int maxDepth(TreeNode *root) {
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// DFS
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if (!root) {
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return 0;
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}
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return max(maxDepth(root->left), maxDepth(root->right)) + 1;
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}
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};
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```
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BFS:
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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* right(right) {}
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* };
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*/
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class Solution {
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public:
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int maxDepth(TreeNode *root) {
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// BFS
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int levels = 0;
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queue<TreeNode *> pending;
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TreeNode *ptr = root;
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if (ptr)
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pending.push(ptr);
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while (!pending.empty()) {
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levels++;
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for (int i = 0, size = pending.size(); i < size; i++) {
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ptr = pending.front();
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pending.pop();
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if (ptr->left)
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pending.push(ptr->left);
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if (ptr->right)
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pending.push(ptr->right);
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}
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}
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return levels;
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}
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};
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```
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