notes/OJ notes/pages/Leetcode Invert-Binary-Tree.md

Leetcode Invert-Binary-Tree

2022-07-06 13:33

Algorithms:

#algorithm #DFS #recursion

Data structures:

#DS #binary_tree

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#recursion tag:#DFS

• Leetcode Insert-Into-a-Binary-Search-Tree
• Leetcode Path-Sum
• Leetcode Symmetric-Tree

Links:

• Link to problem

---

Problem

Given the root of a binary tree, invert the tree, and return its root.

Examples

Example 1:

**Input:** root = [4,2,7,1,3,6,9]
**Output:** [4,7,2,9,6,3,1]

Example 2:

**Input:** root = [2,1,3]
**Output:** [2,3,1]

Example 3:

**Input:** root = []
**Output:** []

Constraints

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

Thoughts

[!summary] This is a #DFS like recursion problem.

Very simple, think of base cases:

• the node is void, skip. And the flow is following

• Catch base case

• Invert sub-trees first

• invert left and right node

Solution

/**
 * 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 {
  void invert(TreeNode *root) {
    if (!root) {
      return;
    }

    invert(root->left);
    invert(root->right);

    TreeNode *tmp = root->left;
    root->left = root->right;
    root->right = tmp;
  }

public:
  TreeNode *invertTree(TreeNode *root) {
    // Using DFS-like Recursion
    invert(root);
    return root;
  }
};