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226. Invert Binary Tree

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

Solutions

Solution 1: Recursion

The idea of recursion is very simple, which is to swap the left and right subtrees of the current node, and then recursively swap the left and right subtrees of the current node.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of nodes in the binary tree.

  • /**
 * 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) {
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return;
            }
            swap(root->left, root->right);
            dfs(root->left);
            dfs(root->right);
        };
        dfs(root);
        return root;
    }
};

  • # 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: Optional[TreeNode]) -> Optional[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.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func invertTree(root *TreeNode) *TreeNode {
	var dfs func(*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.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function invertTree(root: TreeNode | null): TreeNode | null {
    const dfs = (root: TreeNode | null) => {
        if (root === null) {
            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) {
    const dfs = root => {
        if (!root) {
            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.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    #[allow(dead_code)]
    pub fn invert_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        if root.is_none() {
            return root;
        }
        let left = root.as_ref().unwrap().borrow().left.clone();
        let right = root.as_ref().unwrap().borrow().right.clone();
        // Invert the subtree
        let inverted_left = Self::invert_tree(right);
        let inverted_right = Self::invert_tree(left);
        // Update the left & right
        root.as_ref().unwrap().borrow_mut().left = inverted_left;
        root.as_ref().unwrap().borrow_mut().right = inverted_right;
        // Return the root
        root
    }
}

All Problems

All Solutions