leetcode-145-binary-tree-postorder-traversal.swift

/**

Copyright (c) 2020 David Seek, LLC

Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in all
copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM,
DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR
OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE
OR OTHER DEALINGS IN THE SOFTWARE.





https://leetcode.com/problems/binary-tree-postorder-traversal/


Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]
Follow up: Recursive solution is trivial, could you do it iteratively?

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of nodes in root.
Space complexity O(n) where n is the amount of nodes in root.
*/
func inorderTraversal(_ root: TreeNode?) -> [Int] {
        
    var result: [Int] = []
    root?.traversePostOrder { result.append($0) }
    return result
}

extension TreeNode {

    /**
    Will be called recursively 
    until the whole tree is traversed.
    */
    func traversePostOrder(_ visit: (Int) -> Void) {

        // Post order starts at the left branch
        left?.traversePostOrder(visit)

        // Then moves to the right branch
        right?.traversePostOrder(visit)

        // And lastly visits the current node
        visit(val)
    }
}