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题目：
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1
/ 2 5
/ \ 3 4 6
The flattened tree should look like:
1
2
3
4
5
6
代码：
/** * 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: void flatten(TreeNode* root) { if (!root) return; stack<TreeNode *> sta; TreeNode *dummy = new TreeNode(INT_MIN); TreeNode *pre = dummy; dummy->right = root; sta.push(root); while ( !sta.empty() ) { TreeNode *tmp = sta.top(); sta.pop(); if ( tmp->right ) sta.push(tmp->right); if ( tmp->left ) sta.push(tmp->left); tmp->left = NULL; pre->right = tmp; pre = tmp; } } };
tips：
先序遍历（node->left->right）；每次出栈的元素都切断left（为什么right不用切？因为在下一次迭代的时候，pre->right自然就把right的值覆盖了）。
设立一个pre节点，保存上一次出栈的元素。
pre->right = tmp就能按照题意把原有的tree给flatten了。
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学习了一个递归版的代码
/** * 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: void flatten(TreeNode* root) { // terminal condition if (!root) return; // recersive left and child flatten(root->left); flatten(root->right); // if left is not null then do the reconnection if (!root->left) return; TreeNode *p = root->left; while ( p->right) p = p->right; p->right = root->right; root->right = root->left; root->left = NULL; } };
【Flatten Binary Tree to Linked List】cpp
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4505174.html
