翻转二叉树

226. 翻转二叉树

已解答

简单

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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:

翻转二叉树_第1张图片

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

示例 2:

翻转二叉树_第2张图片

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

 思路:用递归来做

对每一层进行翻转,交换每一层的左右子树

 C++代码实现:

  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 nullptr;
     }
     TreeNode *temp=root->left;
     root->left=root->right;
     root->right=temp;
     invertTree(root->left);
     invertTree(root->right);
     return root;
    
    }
};

 JAVA代码实现:

 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) {
       if(root==null){
           return null;
       }
       TreeNode temp=root.left;
       root.left=root.right;
       root.right=temp;
       invertTree(root.left);
       invertTree(root.right);
       return root;
    }
}

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