[LeetCode] 226. Invert Binary Tree

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

翻转二叉树。

给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

思路

是一道不会做，会写 homebrew 也枉然的题。题干即是题意。例子如下，即层层遍历，把左右子树交换。
两种思路，分别是层序遍历（BFS）和深度遍历（DFS）。

复杂度

时间O(n)
空间O(n)

代码

BFS

/**
 * 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) {
		// corner case
		if (root == null) {
			return root;
		}

		// normal case
		Queue<TreeNode> queue = new LinkedList<>();
		queue.offer(root);
		while (!queue.isEmpty()) {
			int size = queue.size();
			for (int i = 0; i < size; i++) {
				TreeNode cur = queue.poll();
				TreeNode temp = cur.left;
				cur.left = cur.right;
				cur.right = temp;
				if (cur.left != null) {
					queue.offer(cur.left);
				}
				if (cur.right != null) {
					queue.offer(cur.right);
				}
			}
		}
		return root;
	}
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
	if (!root) return root;
	let queue = [root];
	while (queue.length) {
		let current = queue.shift();
		if (current === null) continue;
		swap(current);
		queue.push(current.left);
		queue.push(current.right);
	}
	return root;
};

var swap = tree => {
	let temp = tree.left;
	tree.left = tree.right;
	tree.right = temp;
	return tree;
};

DFS

Java实现

/**
 * 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) {
        // corner case
        if (root == null) {
            return root;
        }
        helper(root);
        return root;
    }

    private void helper(TreeNode root) {
        if (root == null) {
            return;
        }
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        helper(root.left);
        helper(root.right);
    }
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
	if (root === null) return root;
	let left = invertTree(root.left);
	let right = invertTree(root.right);
	root.left = right;
	root.right = left;
	return root;
};