114. Flatten Binary Tree to Linked List

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题目

–

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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

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- 求解过程

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题意： 将二叉树转为左子树为空的二叉树，很明显，转化后的二叉树为原二叉树的先序遍历序列；

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-

代码：

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class Solution {
public:
void flatten(TreeNode* root) {
if(root == NULL || (root->left==NULL && root->right == NULL))
{
return ;
}
flatten(root->left); //先转换左子树
flatten(root->right);//再转换右子树
TreeNode* tmpright = root->right; //临时保存右子树
root->right = root->left;  //再将转换后左子树赋给右子树
root->left = NULL;  //再把左子树赋值为空树
TreeNode* tr = root;
while(tr->right)  //找到转换后的右子树（原左子树）的最后一个节点
{
tr = tr->right;
}
tr->right = tmpright; //最后将原右子树拼接到找到的最后的节点之后
}
};