leetcode--Construct Binary Tree from Inorder and Postorder Traversal

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;
    }
};

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原文地址:https://www.cnblogs.com/obama/p/3258791.html