Description
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9] Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3] Output: [2,3,1]
Example 3:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Code
Recursive
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(!root)
return nullptr;
auto left_inverted = invertTree(root->left);
auto right_inverted = invertTree(root->right);
root->right = left_inverted;
root->left = right_inverted;
return root;
}
};Iterative
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
stack<TreeNode*> st;
st.push(root);
while(!st.empty()) {
auto p = st.top(); st.pop();
if(p) {
st.push(p->left);
st.push(p->right);
swap(p->left, p->right);
}
}
return root;
}
};