73前序遍历和中序遍历树构造二叉树

最新推荐文章于 2022-05-28 12:54:44 发布

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#二叉树 #遍历 #递归

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本文介绍如何使用前序遍历和中序遍历构造二叉树的方法，通过递归找到根节点并划分左右子树，提供了两种实现代码。

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根据前序遍历和中序遍历树构造二叉树.

给出中序遍历：[1,2,3]和前序遍历：[2,1,3]. 返回如下的树:

  2
 / \
1   3

解题思路：1.先序遍历的第一个节点为根节点。2.在中序遍历中定位到根节点的位置，左半部分即为左子树，右半部分即为右子树。3.利用递归思想重构二叉树。

已AC的两种代码：/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 

class Solution {
    /**
     *@param preorder : A list of integers that preorder traversal of a tree
     *@param inorder : A list of integers that inorder traversal of a tree
     *@return : Root of a tree
     */
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
       if(preorder.size()==0||inorder.size()==0)
        {
        	return NULL;
    	}
   		TreeNode*sroot=new TreeNode(preorder[0]);
   		sroot->left=NULL;
        sroot->right=NULL;
        int pos=-1;
        vector<int>preleft;
        vector<int>preright;
        vector<int>inleft;
        vector<int>inright;
        int sl=0;
        int sr=0;
   		for(int i=0;i<inorder.size();i++)
            {
            if(inorder[i]==preorder[0])
                {
                pos=i;
                break;
            	}
        	}
        for(int i=0;i<inorder.size();i++)
            {
            if(i<pos)
                {
                inleft.push_back(inorder[i]);
                sl++;
            	}
            if(i>pos)
                {
                inright.push_back(inorder[i]);
                sr++;
            	}
        	}
        for(int i=1;i<preorder.size();i++)
            {
            if(sl)
                {
                preleft.push_back(preorder[i]);
                --sl;
                continue;
            	}
            if(sr)
                {
                preright.push_back(preorder[i]);
                --sr;
            	}
        }
        sroot->left=buildTree(preleft,inleft);
        sroot->right=buildTree(preright,inright);
        return sroot; // write your code here
    }
};

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */


class Solution {
    /**
     *@param preorder : A list of integers that preorder traversal of a tree
     *@param inorder : A list of integers that inorder traversal of a tree
     *@return : Root of a tree
     */
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
       if(preorder.size()==0||inorder.size()==0)
        {
        	return NULL;
    	}
   		TreeNode*sroot=new TreeNode(preorder[0]);
   		sroot->left=NULL;
        sroot->right=NULL;
        vector<int>preleft;
        vector<int>preright;
        vector<int>inleft;
        vector<int>inright;
        int pos=find(inorder.begin(),inorder.end(),preorder[0])-inorder.begin();
        inleft.assign(inorder.begin(),inorder.begin()+pos);
        inright.assign(inorder.begin()+pos+1,inorder.end());
        preleft.assign(preorder.begin()+1,preorder.begin()+pos+1);
        preright.assign(preorder.begin()+pos+1,preorder.end());
        sroot->left=buildTree(preleft,inleft);
        sroot->right=buildTree(preright,inright);
        return sroot; // write your code here
    }
};