leetcode 226 Invert Binary Tree 翻转二叉树

流川疯

发布于 2022-05-06 14:00:21

3170

发布于 2022-05-06 14:00:21

大牛没有能做出来的题，我们要好好做一做

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia: This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

递归解决方案：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) 
    {
        if(root ==NULL) return root;
        TreeNode* node = invertTree(root->left);
        root->left = invertTree(root->right);
        root->right = node;
        return root;
    }
};

非递归解决方案：

 TreeNode* invertTree(TreeNode* root) 
    {
        if(root == NULL)return NULL;
        vector<TreeNode*> stack;
        stack.push_back(root);
        while(!stack.empty())
        {
            TreeNode* node = stack.back();// or stack.top()
            stack.pop_back();
            swap(node->left,node->right);
            if(node->left)stack.push_back(node->left);
            if(node->right)stack.push_back(node->right);
        }
        return root;
    }

python：

def invertTree(self, root):
    if root:
        root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
        return root


Maybe make it four lines for better readability:

def invertTree(self, root):
    if root:
        invert = self.invertTree
        root.left, root.right = invert(root.right), invert(root.left)
        return root


--------------------------------------------------------------------------------

And an iterative version using my own stack:

def invertTree(self, root):
    stack = [root]
    while stack:
        node = stack.pop()
        if node:
            node.left, node.right = node.right, node.left
            stack += node.left, node.right
    return root

def invertTree(self, root):
    if root is None:
        return None
    root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
    return root

python非递归解决方案：

DFS version:

def invertTree(self, root):
        if (root):
            self.invertTree(root.left)
            self.invertTree(root.right)
            root.left, root.right = root.right, root.left
            return root   


BFS version:

def bfs_invertTree(self, root):
        queue = collections.deque()
        if (root):
            queue.append(root)

        while(queue):
            node = queue.popleft()
            if (node.left):
                queue.append(node.left)
            if (node.right):
                queue.append(node.right)
            node.left, node.right = node.right, node.left

        return root

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原始发表：2015-06-23，如有侵权请联系 cloudcommunity@tencent.com 删除

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