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  • [LintCode] Binary Tree Flipping

    Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

    Example

    Given a binary tree {1,2,3,4,5}

        1
       / 
      2   3
     / 
    4   5
    

    return the root of the binary tree {4,5,2,#,#,3,1}.

       4
      / 
     5   2
        / 
       3   1  


    Think Out Loud


    For a given node n whose left child is not null, we do the following changes.

    n.left.left = n.right;

    n.left.right = n;

    n.left = null;

    n.right = null.

    The new root after flipping is the leftmost node in the original tree.

    We can solve this recursively bottom up. Solving it top down is more complicated because we need the top nodes to access 

    the down nodes. Changing the top nodes makes this process a lot harder.

    O(n) runtime, O(H) space, H is the max depth of the give binary tree.



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  • 原文地址:https://www.cnblogs.com/lz87/p/7478954.html
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