LeetCode_Construct Binary Tree from Inorder and Postorder Traversal

一.題目

Construct Binary Tree from Inorder and Postorder Traversal

  Total Accepted: 33418 Total Submissions: 124726My Submissions

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

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

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二.解題技巧

    這道題和Construct Binary Tree from Preorder and Inorder Traversal類似，都是考察基本概念的，後序遍歷是先遍歷左子樹，然後遍歷右子樹，最後遍歷根節點。

    做法都是先根據後序遍歷的概念，找到後序遍歷最後的一個值，即爲根節點的值，然後根據根節點將中序遍歷的結果分成左子樹和右子樹，然後就可以遞歸的實現了。

    上述做法的時間複雜度爲O(n^2)，空間複雜度爲O(1)。

    

三.實現代碼

#include <iostream>
#include <algorithm>
#include <vector>


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

using std::vector;
using std::find;

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

class Solution
{
private:
    TreeNode* buildTree(vector<int>::iterator PostBegin, vector<int>::iterator PostEnd,
                        vector<int>::iterator InBegin, vector<int>::iterator InEnd)
    {
        if (InBegin == InEnd)
        {
            return NULL;
        }

        if (PostBegin == PostEnd)
        {
            return NULL;
        }

        int HeadValue = *(--PostEnd);
        TreeNode *HeadNode = new TreeNode(HeadValue);

        vector<int>::iterator LeftEnd = find(InBegin, InEnd, HeadValue);
        if (LeftEnd != InEnd)
        {
            HeadNode->left = buildTree(PostBegin, PostBegin + (LeftEnd - InBegin),
                             InBegin, LeftEnd);
        }

        HeadNode->right = buildTree(PostBegin + (LeftEnd - InBegin), PostEnd,
                                LeftEnd + 1, InEnd);

        return HeadNode;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)
    {
        if (inorder.empty())
        {
            return NULL;
        }

        return buildTree(postorder.begin(), postorder.end(), inorder.begin(),
                         inorder.end());

    }
};

四.體會

   這道題主要考察的是基本概念，並沒有很複雜的算法在裏面，可以算是對於二叉樹的遍歷的進一步理解。

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