105 - Construct Binary Tree from Preorder and Inorder Traversal

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

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

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思路分析：

题意：也就是根据树的前序中序遍历来重构二叉树。

There is an example.

        _______7______
       /              \
    __10__          ___2
   /      \        /
   4       3      _8
            \    /
             1  11

The preorder and inorder traversals for the binary tree above is:

preorder = {7,10,4,3,1,2,8,11}
inorder = {4,10,3,1,7,11,8,2}

The first node in preorder alwasy the root of the tree. We can break the tree like:
1st round:
preorder:  {7}, {10,4,3,1}, {2,8,11}
inorder:     {4,10,3,1}, {7}, {11, 8,2}

        _______7______
       /              \
    {4,10,3,1}       {11,8,2}

Since we alreay find that {7} will be the root, and in "inorder" sert, all the data in the left of {7} will construct the left sub-tree. And the right part will construct a right sub-tree. We can the left and right part agin based on the preorder.

2nd round
left part                                                                            right part
preorder: {10}, {4}, {3,1}                                              {2}, {8,11}
inorder:  {4}, {10}, {3,1}                                                {11,8}, {2}

        _______7______
       /              \
    __10__          ___2
   /      \        /
   4      {3,1}   {11,8}

see that, {10} will be the root of left-sub-tree and {2} will be the root of right-sub-tree.

Same way to split {3,1} and {11,8}, yo will get the complete tree now.

        _______7______
       /              \
    __10__          ___2
   /      \        /
   4       3      _8
            \    /
             1  11

So, simulate this process from bottom to top with recursion as following code.

代码如下：

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

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

class Solution_105_ConstructBinaryTreefromPreorderandInorderTraversal
{
public:
	TreeNode *BuildTreePI(vector<int> &preorder, vector<int> &inorder, int p_s, int p_e, int i_s, int i_e)
	{
		


		int pivot = preorder[p_s];
		int i = i_s;

		for (; i < i_e; i++)
		{
			if (inorder[i] == pivot)
			{
				break;
			}
		}

		int length1 = i - i_s - 1; //中序遍历中的root在向量中的下标
		int length2 = i_e - i - 1; //中序遍历中从后向前的元素个数
		TreeNode *node = new TreeNode(pivot);

		node->left = BuildTreePI(preorder, inorder, p_s + 1, length1 + p_s + 1, i_s, i - 1);
		node->right = BuildTreePI(preorder, inorder, p_e - length2, p_e, i + 1, i_e);
	}
	TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) 
	{
		return BuildTreePI(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
	}
};