77 lines
1.6 KiB
C++
77 lines
1.6 KiB
C++
// Source : https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/
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// Author : Hao Chen
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// Date : 2014-07-03
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/**********************************************************************************
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*
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* Given a binary tree, flatten it to a linked list in-place.
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*
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* For example,
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* Given
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*
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* 1
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* / \
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* 2 5
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* / \ \
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* 3 4 6
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*
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* The flattened tree should look like:
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*
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* 1
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* \
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* 2
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* \
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* 3
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* \
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* 4
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* \
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* 5
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* \
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* 6
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*
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*
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* Hints:
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* If you notice carefully in the flattened tree, each node's right child points to
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* the next node of a pre-order traversal.
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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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void flatten(TreeNode *root) {
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vector<TreeNode*> v, stack;
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stack.push_back(root);
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while(stack.size()>0){
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TreeNode* node = stack.back();
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stack.pop_back();
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v.push_back(node);
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if (node && node->right){
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stack.push_back(node->right);
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}
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if (node && node->left){
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stack.push_back(node->left);
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}
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}
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v.push_back(NULL);
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for(int i=0; i<v.size(); i++){
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if (v[i]){
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v[i]->left = NULL;
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v[i]->right = v[i+1];
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}
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}
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}
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};
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