Invert Binary Tree

 

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

to

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

 

Analyse: the same as Binary Tree Level Order Traversal except that we need to swap the left and right subtrees of the current node.

1. Recursion: First swap the left and right subtree of the roots, then recursively swap the left and right subtrees of them...

    Runtime: 0ms.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(!root) return NULL;
        
        TreeNode* leftSub = root->left; //swap the left and right subtree of the root
        root->left = root->right;
        root->right = leftSub;
        invertTree(root->left);
        invertTree(root->right);
        
        return root;
    }
};

 

2. Iteration: swap the left and right subtree of the current node and push the new left and right subtree into a queue. 

    Runtime: 0ms.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(!root) return NULL;
        
        queue<TreeNode* > qu;
        qu.push(root);
        while(!qu.empty()){
            int n = qu.size();
            while(n--){
                TreeNode* temp = qu.front();
                qu.pop();
                TreeNode* tempLeft = temp->left; //swap the left and right subtree of the current root
                temp->left = temp->right;
                temp->right = tempLeft;
                
                if(temp->left) qu.push(temp->left); //push the original right subtree of the root first
                if(temp->right) qu.push(temp->right); //continuously do the process until all nodes are visited
            }
        }
        return root;
    }
};