226. Invert Binary Tree

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problem

solution

option 1 - backtracking

  • inorder traverse version
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class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(!root) return root;
        // post-order
        
        root->left = invertTree(root->left);
        root->right = invertTree(root->right);

        TreeNode* temp = root->left;
        root->left = root->right;
        root->right = temp;
        
        return root;
    }
};
  • postorder traverse version
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class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(!root) return root;
        
        root->left = invertTree(root->left);
        root->right = invertTree(root->right);
        TreeNode* temp = root->left;
        root->left = root->right;
        root->right = temp;
        
        return root;
    }
};

option 2 - bfs

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class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if(!root) return root;
        queue<TreeNode * >q;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            while(size-- > 0){
                TreeNode *p = q.front();
                q.pop();
                if(p->left) q.push(p->left);
                if(p->right) q.push(p->right);
                TreeNode *temp = p->left;
                p->left = p->right;
                p->right = temp;
            }
        }
        return root;
    }
};

analysis

  • time complexity O(n)
  • space complexity O(n)