Code & Func

第41天。

Invert a binary tree.

1
     4
2
   /   \
3
  2     7
4
 / \   / \
5
1   3 6   9

to

1
     4
2
   /   \
3
  7     2
4
 / \   / \
5
9   6 3   1

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.

emmmm，挺出名的一道题目。

其实挺简单：

1
TreeNode* invertTre1e(TreeNode* root) {
2
    if (root == nullptr) return root;
3
    TreeNode *left = invertTree(root->right);
4
    TreeNode *right = invertTree(root->left);
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    root->left = left;
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    root->right = right;
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    return root;
8
}

然后是迭代的方法:

1
TreeNode* invertTree(TreeNode *root) {
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    stack<TreeNode *> st;
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    st.push(root);
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    TreeNode *ret =root;
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    while(!st.empty()) {
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        root = st.top();
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        st.pop();
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        if (root) {
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            st.push(root->left);
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            st.push(root->right);
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            TreeNode *t = root->left;
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            root->left = root->right;
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            root->right = t;
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        }
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    }
2 collapsed lines
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    return ret;
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}