226. 翻转二叉树

题目描述

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。

 

示例 1：

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]

示例 2：

输入：root = [2,1,3]
输出：[2,3,1]

示例 3：

输入：root = []
输出：[]

 

提示：

• 树中节点数目范围在 [0, 100] 内
• -100 <= Node.val <= 100

方法一：递归

递归的思路很简单，就是交换当前节点的左右子树，然后递归地交换当前节点的左右子树。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点个数。

Java

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);
    }
}

C++

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;
    }
};

TypeScript

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;
}

Python

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

方法二

Java

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode l = invertTree(root.left);
        TreeNode r = invertTree(root.right);
        root.left = r;
        root.right = l;
        return root;
    }
}

C++

class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) {
            return root;
        }
        TreeNode* l = invertTree(root->left);
        TreeNode* r = invertTree(root->right);
        root->left = r;
        root->right = l;
        return root;
    }
};

TypeScript

function invertTree(root: TreeNode | null): TreeNode | null {
    if (!root) {
        return root;
    }
    const l = invertTree(root.left);
    const r = invertTree(root.right);
    root.left = r;
    root.right = l;
    return root;
}

Python

class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is None:
            return None
        l, r = self.invertTree(root.left), self.invertTree(root.right)
        root.left, root.right = r, l
        return root