Invert a binary tree.
4 / \ 2 7 / \ / \ 1 3 6 9to
4 / \ 7 2 / \ / \ 9 6 3 11.BFS
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root == NULL) return NULL;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()){
TreeNode* node = q.front();
q.pop();
TreeNode* node2 = node->left;
node->left = node->right;
node->right = node2;
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
return root;
}
};
2.DFS
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root == NULL) return root;
TreeNode* node = root->left;
root->left = root->right;
root->right = node;
if(root->left) invertTree(root->left);
if(root->right) invertTree(root->right);
return root;
}
};