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[LeetCode]*106.Construct Binary Tree from Inorder and Postorder Traversal

来源：程序员人生发布时间：2015-06-18 08:54:53 阅读次数：3679次

题目

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

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

思路

思路和[LeetCode]*105.Construct Binary Tree from Preorder and Inorder Traversal1样。

代码

/*---------------------------------------
*   日期：2015-05-01
*   作者：SJF0115
*   题目: 106.Construct Binary Tree from Inorder and Postorder Traversal
*   网址：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
*   结果：AC
*   来源：LeetCode
*   博客：
-----------------------------------------*/
#include <iostream>
#include <vector>
using namespace std;

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

class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        int size = inorder.size();
        if(size <= 0){
            return nullptr;
        }//if
        return InPostBuildTree(inorder,postorder,0,size-1,size);
    }
private:
    TreeNode* InPostBuildTree(vector<int> &inorder,vector<int> &postorder,int inIndex,int postIndex,int size){
        if(size <= 0){
            return nullptr;
        }//if
        // 根节点
        TreeNode* root = new TreeNode(postorder[postIndex]);
        // 寻觅postorder[postIndex]在中序序列中的下标
        int index = 0;
        for(int i = 0;i < size;++i){
            if(postorder[postIndex] == inorder[inIndex+i]){
                index = inIndex+i;
                break;
            }//if
        }//for
        int leftSize = index - inIndex;
        int rightSize = size - leftSize - 1;
        root->left = InPostBuildTree(inorder,postorder,inIndex,postIndex-1-rightSize,leftSize);
        root->right = InPostBuildTree(inorder,postorder,index+1,postIndex-1,rightSize);
        return root;
    }
};

void PreOrder(TreeNode* root){
    if(root){
        cout<<root->val<<endl;
        PreOrder(root->left);
        PreOrder(root->right);
    }//if
}

int main() {
    Solution solution;
    vector<int> inorder = {8,4,2,5,1,6,3,7};
    vector<int> postorder = {8,4,5,2,6,7,3,1};
    TreeNode* root = solution.buildTree(inorder,postorder);
    PreOrder(root);
}

运行时间

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