226. Invert Binary Tree

O(N) Time recursive solution

• if root is null return null
• we just need to swap the left and right children of each node recursively.we can use inbuilt swap function or implement our own swap function.
• We can travel in preorder as well as postorder , both solutions are accepted.(here is preorder solution)

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(!root) return NULL;

        // Swap left and right children
        // swap(node->left, node->right);
        TreeNode* temp;
        temp = root->left;
        root->left = root->right;
        root->right = temp;

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

        return root;
    }
};

O(N) Time O(N) stack iterative solution

• We use stack instead of recursive stack.

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        stack<TreeNode*> st;
        st.push(root);

        while (!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            if (node) {
                st.push(node->left);
                st.push(node->right);
                swap(node->left, node->right);
            }
        }
        return root;
    }
};

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Many might say why we need an iterative solution in every tree problem ?

The ans is simple, Tree can be skewed which effectively makes it a linked list. And if we recurse in a large linked-list, the call stack will overflow for sure. That’s why its necessary to also know how to implement iteratively. Honestly, iterative implementation also will utilize a stack, but it will be in heap.