LeetCode_Construct Binary Tree from Inorder and Postorder Traversal

一.题目

Construct Binary Tree from Inorder and Postorder Traversal

   Total Accepted: 33418  Total Submissions: 124726 My 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.解题技巧

    这道题和Construct Binary Tree from Preorder and Inorder Traversal类似，都是考察基本概念的，后序遍历是先遍历左子树，然后遍历右子树，最后遍历根节点。

    做法都是先根据后序遍历的概念，找到后序遍历最后的一个值，即为根节点的值，然后根据根节点将中序遍历的结果分成左子树和右子树，然后就可以递归的实现了。

    上述做法的时间复杂度为O(n^2)，空间复杂度为O(1)。

    

三.实现代码

#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.体会

   这道题主要考察的是基本概念，并没有很复杂的算法在里面，可以算是对于二叉树的遍历的进一步理解。

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