Question

226. 翻转二叉树

English Version

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

解法

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。

# 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.
 * 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.
 * 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.
// #[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
  }
}

/**
 * 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;
};

方法二

# 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]:
    if root is None:
      return None
    l, r = self.invertTree(root.left), self.invertTree(root.right)
    root.left, root.right = r, l
    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) {
    if (root == null) {
      return null;
    }
    TreeNode l = invertTree(root.left);
    TreeNode r = invertTree(root.right);
    root.left = r;
    root.right = l;
    return root;
  }
}

/**
 * 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) {
    if (!root) {
      return root;
    }
    TreeNode* l = invertTree(root->left);
    TreeNode* r = invertTree(root->right);
    root->left = r;
    root->right = l;
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func invertTree(root *TreeNode) *TreeNode {
  if root == nil {
    return root
  }
  l, r := invertTree(root.Left), invertTree(root.Right)
  root.Left, root.Right = r, l
  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 {
  if (!root) {
    return root;
  }
  const l = invertTree(root.left);
  const r = invertTree(root.right);
  root.left = r;
  root.right = l;
  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) {
  if (!root) {
    return root;
  }
  const l = invertTree(root.left);
  const r = invertTree(root.right);
  root.left = r;
  root.right = l;
  return root;
};