我们都知道，已知中序和后序的序列是可以唯一确定一个二叉树的。

初始化时候二叉树为：==================

中序遍历序列，           ======O===========

后序遍历序列，           =================O

红色部分是左子树，黑色部分是右子树，O是根节点

如上图所示，O是根节点，由后序遍历可知，

根据这个O可以把找到其在中序遍历当中的位置，进而，知道当前这个根节点O的左子树的前序遍历和中序遍历序列的范围。

以及右子树的前序遍历和中序遍历序列的范围。

到这里返现出现了重复的子问题，而且子问题的规模没有原先的问题大，即红色部分和黑色部分。

而联系这两个子问题和原先的大问题的纽带是这个找到的根节点。

可以选择用递归来解决这个问题，递归的结束条件是子问题序列里面只有一个元素。

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

给定一个二叉树的中序和后序遍历序列，构造这个二叉树。

笔记：

你可以假定，这棵树里面没有重复的节点。

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

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

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

 
 
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) {}
 * };
 */
TreeNode *build(vector<int> &inorder, int left1, int right1, vector<int> &postorder, int left2, int right2)
{

if(right1 - left1 != right2 - left2)
    {
        return NULL;
    }
    if(right1 >= inorder.size() || right2 >= postorder.size())
    {
        return NULL;
    }

//递归结束条件
    if(left1 == right1 && left2 == right2)
    {
        TreeNode *root = new TreeNode(inorder[left1]);
        return root;
    }
    else if(left1 < right1 && left2 < right2)
    {

TreeNode *root = new TreeNode(postorder[right2]);
        int i;
        for(i = right1; i >= left1; i--)
        {
            //找到中序遍历的根节点
            if(inorder[i] == postorder[right2])
            {
                break;
            }
        }
        if(i < left1)
        {
            return NULL;
        }
        root->left = build(inorder, left1, i - 1, postorder, left2, right2 + i - right1 - 1);
        root->right = build(inorder, i + 1, right1, postorder, right2 + i - right1, right2 - 1);
        return root;

}
    else
    {
        return NULL;
    }

}

TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
{
    return build(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
}

vector<vector<int> > levelOrder(TreeNode *root)
{

vector<vector<int> > matrix;
    if(root == NULL)
    {
        return matrix;
    }
    vector<int> temp;
    temp.push_back(root->val);
    matrix.push_back(temp);

vector<TreeNode *> path;
    path.push_back(root);

int count = 1;
    while(!path.empty())
    {
        TreeNode *tn = path.front();
        if(tn->left)
        {
            path.push_back(tn->left);
        }
        if(tn->right)
        {
            path.push_back(tn->right);
        }
        path.erase(path.begin());
        count--;

if(count == 0)
        {
            vector<int> tmp;
            vector<TreeNode *>::iterator it = path.begin();
            for(; it != path.end(); ++it)
            {
                tmp.push_back((*it)->val);
            }
            if(tmp.size() > 0)
            {
                matrix.push_back(tmp);
            }
            count = path.size();
        }
    }
    return matrix;
}

// 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);

int in[7] = {1, 7, 3, 4, 5, 6, 2};
    int post[7] = {1, 3, 5, 4, 7, 2, 6};
    vector<int> inorder(in, in + 7), postorder(post, post + 7);

TreeNode *root = buildTree(inorder, postorder);

vector<vector<int> > ans = levelOrder(root);

for (int i = 0; i < ans.size(); ++i)
    {
        for (int j = 0; j < ans[i].size(); ++j)
        {
            cout << ans[i][j] << " ";
        }
        cout << endl;
    }
    DestroyTree(root);
    return 0;
}

 
结果输出：
6

7 2

1 4

3 5

ps.测试的输出用的是层次遍历
 
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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