114 lines
2 KiB
Markdown
114 lines
2 KiB
Markdown
# Leetcode Invert-Binary-Tree
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#### 2022-07-06 13:33
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> ##### Algorithms:
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> #algorithm #DFS #recursion
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> ##### Data structures:
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> #DS #binary_tree
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#recursion tag:#DFS
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```
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- [[Leetcode Insert-Into-a-Binary-Search-Tree]]
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- [[Leetcode Path-Sum]]
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- [[Leetcode Symmetric-Tree]]
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##### Links:
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- [Link to problem](https://leetcode.com/problems/invert-binary-tree/)
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___
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### Problem
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Given the `root` of a binary tree, invert the tree, and return _its root_.
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#### Examples
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**Example 1:**
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```
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**Input:** root = [4,2,7,1,3,6,9]
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**Output:** [4,7,2,9,6,3,1]
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```
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**Example 2:**
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```
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**Input:** root = [2,1,3]
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**Output:** [2,3,1]
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```
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**Example 3:**
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```
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**Input:** root = []
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**Output:** []
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```
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#### Constraints
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- The number of nodes in the tree is in the range `[0, 100]`.
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- `-100 <= Node.val <= 100`
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### Thoughts
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> [!summary]
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> This is a #DFS like recursion problem.
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Very simple, think of base cases:
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- the node is void, skip.
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And the flow is following
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- Catch base case
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- Invert sub-trees first
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- invert left and right node
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### Solution
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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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void invert(TreeNode *root) {
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if (!root) {
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return;
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}
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invert(root->left);
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invert(root->right);
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TreeNode *tmp = root->left;
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root->left = root->right;
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root->right = tmp;
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}
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public:
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TreeNode *invertTree(TreeNode *root) {
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// Using DFS-like Recursion
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invert(root);
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return root;
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}
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};
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``` |