题目链接

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

题意

由二叉树的先序遍历和中序遍历建树

思路

理解建树过程;使用递归,递归关键:清楚递归过程,明确函数参数、返回值,写终止条件。

此外,注意空树的特判。

其他点

1 熟悉find()、assign()的使用。

2 使用

struct TreeNode *pNode =new TreeNode(*preRootIter);

而不是

struct TreeNode node=TreeNode(*preRootIter);
struct TreeNode* pNode=node;

否则造成assign的时候树中某些节点值被改变。

todo

上述问题原因待查。

应该最后要遍历树释放内存吧。

代码

#include <vector>
#include <iostream>
using namespace std; struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}; class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if(!preorder.size()){
return NULL;
}
else{
struct TreeNode* root=buildT(preorder,inorder);
return root;
}
} private:
TreeNode* buildT(vector<int> preorder,vector<int> inorder){
vector<int>::iterator preRootIter=preorder.begin();
vector<int>::iterator midRootIter=find(inorder.begin(), inorder.end(), *preRootIter);
//create a new node
struct TreeNode *pNode =new TreeNode(*preRootIter);
// struct TreeNode node=TreeNode(*preRootIter);
// struct TreeNode* pNode=node; //left child tree
vector<int>::size_type leftLen=midRootIter-inorder.begin();
vector<int>::iterator lPreBeg=preRootIter+1;
vector<int>::iterator lPreEnd=lPreBeg+leftLen;
if(!leftLen){
pNode->left=NULL;
}
else{
vector<int> preorderLeft;
preorderLeft.assign(lPreBeg, lPreEnd); vector<int> inorderLeft;
inorderLeft.assign(inorder.begin(), midRootIter);
pNode->left=buildT(preorderLeft, inorderLeft);
} //right child tree
vector<int>::size_type rightLen=inorder.end()-(midRootIter+1);
if(!rightLen){
pNode->right=NULL;
}
else{
vector<int> preorderRight;
vector<int>::iterator rPreBeg=lPreEnd;
vector<int>::iterator rPreEnd=rPreBeg+rightLen;
preorderRight.assign(rPreBeg, rPreEnd); vector<int> inorderRight;
inorderRight.assign(midRootIter+1, inorder.end());
pNode->right=buildT(preorderRight, inorderRight);
} return pNode;
}
}; int main(){
Solution solution;
vector<int> preorder={3,9,20,15,7};
vector<int> inorder={9,3,15,20,7};
struct TreeNode* tree=solution.buildTree(preorder,inorder);
return 0;
}

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