Question

Formatted question description: https://leetcode.ca/all/226.html

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

Algorithm

Exchange the current left and right nodes, and call recursion directly.

Code

import java.util.Stack;

public class Invert_Binary_Tree {
    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode invertTree(TreeNode root) {

            if (root == null) {
                return root;
            }

            // bfs
            Stack<TreeNode> sk = new Stack<>();
            sk.push(root);

            while (!sk.isEmpty()) {
                TreeNode current = sk.pop();

                if (current.left != null) {
                    sk.push(current.left);
                }

                if (current.right != null) {
                    sk.push(current.right);
                }

                // switch
                TreeNode tmp;
                tmp = current.left;
                current.left = current.right;
                current.right = tmp;
            }

            return root;

        }
    }

    class Solution_recursion {
        public TreeNode invertTree(TreeNode root) {
            if (root == null) {
                return root;
            }

            // swap
            TreeNode tmp = root.left;
            root.left = root.right;
            root.right = tmp;

            invertTree(root.left);
            invertTree(root.right);

            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);
    }
}

// OJ: https://leetcode.com/problems/invert-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) return NULL;
        swap(root->left, root->right);
        invertTree(root->left);
        invertTree(root->right);
        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: 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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def invertTree(self, root):
    """
    :type root: TreeNode
    :rtype: TreeNode
    """
    if not root:
      return
    root.left, root.right = root.right, root.left
    self.invertTree(root.left)
    self.invertTree(root.right)
    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(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;
};

/**
 * 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;
}

226 - Invert Binary Tree

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