本质上是二叉树的root->right->left遍历。

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

给定一个二叉树,就地的把他转换成一个链表。

例如:

给定

         1
/ \
2 5
/ \ \
3 4 6

转换后的树应该向这样子:
   1
\
2
\
3
\
4
\
5
\
6

提示:

如果你观察的足够仔细,你会发现,每个还在的右孩子指针指向了,这个节点在前序遍历中的后面的那个节点。

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

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
Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
迭代版本:
test.cpp:
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#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 * int val;
 * TreeNode *left;
 * TreeNode *right;
 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
void flatten(TreeNode *root)
{
    if (root == NULL)
    {
        return ;
    }
    stack<TreeNode *> s;
    s.push(root);

TreeNode *tmp;
    while (!s.empty())
    {
        tmp = s.top();
        s.pop();

if (tmp->right)
        {
            s.push(tmp->right);
        }
        if (tmp->left)
        {
            s.push(tmp->left);
        }

tmp->left = NULL;
        if (!s.empty())
        {
            tmp->right = s.top();
        }
    }
}

// 树中结点含有分叉,
//                  6
//              /       \
//             7         2
//           /   \
//          1     4
//               / \
//              3   5
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(7);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(2);
    TreeNode *pNodeA4 = CreateBinaryTreeNode(1);
    TreeNode *pNodeA5 = CreateBinaryTreeNode(4);
    TreeNode *pNodeA6 = CreateBinaryTreeNode(3);
    TreeNode *pNodeA7 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
    ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
    ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

flatten(pNodeA1);

TreeNode *trav = pNodeA1;
    while (trav != NULL)
    {
        cout << trav->val << " ";
        trav = trav->right;
    }
    cout << endl;

DestroyTree(pNodeA1);
    return 0;
}

 
 
递归版本:
test.cpp:
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#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 * int val;
 * TreeNode *left;
 * TreeNode *right;
 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
void flatten(TreeNode *root)
{

if(root == NULL)
    {
        return ;
    }
    if(root->left == NULL && root->right == NULL)
    {
        return ;
    }

flatten(root->left);
    flatten(root->right);

TreeNode *tmpright = root->right;
    /*因为是前序遍历*/
    root->right = root->left;
    root->left = NULL;
    TreeNode *tmp = root;
    while(tmp->right)
    {
        /*找到右子树的前序遍历的最后一个节点*/
        tmp = tmp->right;
    }
    tmp->right = tmpright;

return ;
}

// 树中结点含有分叉,
//                  6
//              /       \
//             7         2
//           /   \
//          1     4
//               / \
//              3   5
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(7);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(2);
    TreeNode *pNodeA4 = CreateBinaryTreeNode(1);
    TreeNode *pNodeA5 = CreateBinaryTreeNode(4);
    TreeNode *pNodeA6 = CreateBinaryTreeNode(3);
    TreeNode *pNodeA7 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
    ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
    ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

flatten(pNodeA1);

TreeNode *trav = pNodeA1;
    while (trav != NULL)
    {
        cout << trav->val << " ";
        trav = trav->right;
    }
    cout << endl;

DestroyTree(pNodeA1);
    return 0;
}

结果输出:
6 7 1 4 3 5 2
BinaryTree.h:
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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *CreateBinaryTreeNode(int value);
void ConnectTreeNodes(TreeNode *pParent,
                      TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.cpp:
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#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

//创建结点
TreeNode *CreateBinaryTreeNode(int value)
{
    TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->left = pLeft;
        pParent->right = pRight;
    }
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
            printf("value of its left child is: %d.\n", pNode->left->val);
        else
            printf("left child is null.\n");

if(pNode->right != NULL)
            printf("value of its right child is: %d.\n", pNode->right->val);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

printf("\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
    PrintTreeNode(pRoot);

if(pRoot != NULL)
    {
        if(pRoot->left != NULL)
            PrintTree(pRoot->left);

if(pRoot->right != NULL)
            PrintTree(pRoot->right);
    }
}

void DestroyTree(TreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        TreeNode *pLeft = pRoot->left;
        TreeNode *pRight = pRoot->right;

delete pRoot;
        pRoot = NULL;

DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}

 

 

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