leetcode--Construct Binary Tree from Inorder and Postorder Traversal-优快云博客

本文详细介绍了如何通过中序和后序遍历序列构建二叉树，强调了数组中元素不重复的重要性。提供了一个C++类实现的解决方案，并通过实例演示了解题过程。

1.题目描述

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

Note:

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

2.解法分析

理解了什么是后序，什么是中序，此题迎刃而解，需要注意的是，解此题需要数组中没有相同的元素，不然的话可能性就多了。

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

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

 * };

*/

class Solution {

public:

    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        if(inorder.size()!=postorder.size()||inorder.size()==0)return NULL;

        return myBuildTree(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);

    TreeNode *myBuildTree(vector<int> &inorder, vector<int> &postorder,int iStart,int iEnd,int pStart,int pEnd){

        if(iStart>iEnd)return NULL;

        TreeNode *parent = new TreeNode(postorder[pEnd]);

        int rootIndex = iStart;

        while(inorder[rootIndex]!=postorder[pEnd]){rootIndex++;}

        parent->left = myBuildTree(inorder,postorder,iStart,rootIndex-1,pStart,pStart+rootIndex-iStart-1);

        parent->right = myBuildTree(inorder,postorder,rootIndex+1,iEnd,pEnd-(iEnd-rootIndex),pEnd-1);

        return parent;

};

转载于:https://www.cnblogs.com/obama/p/3258791.html