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
DFS.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
def dfs(root):
if root is None:
return
root.left, root.right = root.right, root.left
dfs(root.left)
dfs(root.right)
dfs(root)
return root
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
dfs(root);
return root;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
TreeNode t = root.left;
root.left = root.right;
root.right = t;
dfs(root.left);
dfs(root.right);
}
}
/**
* 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) {
dfs(root);
return root;
}
void dfs(TreeNode* root) {
if (!root) return;
TreeNode* t = root->left;
root->left = root->right;
root->right = t;
dfs(root->left);
dfs(root->right);
}
};
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func invertTree(root *TreeNode) *TreeNode {
var dfs func(root *TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
root.Left, root.Right = root.Right, root.Left
dfs(root.Left)
dfs(root.Right)
}
dfs(root)
return root
}
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {TreeNode}
*/
var invertTree = function (root) {
function dfs(root) {
if (!root) return;
[root.left, root.right] = [root.right, root.left];
dfs(root.left);
dfs(root.right);
}
dfs(root);
return root;
};