#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/graph-theory/tree-fundamentals/invert-binary-tree-implementation

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/graph-theory/tree-fundamentals/invert-binary-tree-implementation

Repovive

LeetCode 226 Invert Binary Tree - Implementation - Tree Fundamentals | Graph Theory | Repovive

Repovive.

Graph Theory37 sections · 1633 units81h total

Open in Course

LeetCode 226 Invert Binary Tree - Implementation

Code solution

Here is the solution:

function invertTree(root):
    if root is null:
        return null

    temp = root.left
    root.left = root.right
    root.right = temp

    invertTree(root.left)
    invertTree(root.right)

    return root

You swap the pointers with a temp variable. Then you recurse on the swapped children. The recursion propagates the inversion down the tree. Finally, you return the root.

This runs in $O(n)$ time and uses $O(h)$ space.