70 lines
1.8 KiB
C++
70 lines
1.8 KiB
C++
// Source : https://oj.leetcode.com/problems/binary-tree-upside-down/
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// Author : Hao Chen
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// Date : 2014-11-17
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/**********************************************************************************
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* Given a binary tree where all the right nodes are either leaf nodes with
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* a sibling (a left node that shares the same parent node) or empty,
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*
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* Flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes.
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* Return the new root.
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*
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* For example:
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* Given a binary tree {1,2,3,4,5},
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* 1
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* / \
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* 2 3
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* / \
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* 4 5
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* return the root of the binary tree [4,5,2,#,#,3,1].
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* 4
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* / \
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* 5 2
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* / \
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* 3 1
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* confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
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*
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**********************************************************************************/
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/**
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* Definition for binary tree
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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TreeNode *upsideDownBinaryTree(TreeNode *root) {
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//using a dummy node to help to store the new tree
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TreeNode dummy(0);
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TreeNode *head = &dummy, *left=NULL, *right=NULL;
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while ( root!=NULL ) {
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//find the right & left
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left = root->right;
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right = root;
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//move root the next
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root = root->left;
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//replace the right with current root
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right->left = head->left;
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right->right = head->right;
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//move the dummy to the root
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dummy.right = right;
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dummy.left = left;
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//reset the head to the root
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head = &dummy;
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}
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return head->right;
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}
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};
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