输入某二叉树的前序遍历和中序遍历的结果，重建出该二叉树

2024年6月9日 123次阅读

题目描述

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

思路

递归建立左子树和右子树

代码

java：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
         TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);
        return root;
    }
    //前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}
    private TreeNode reConstructBinaryTree(int [] pre,int startPre,int endPre,int [] in,int startIn,int endIn) {
          
        if(startPre>endPre||startIn>endIn)
            return null;
        TreeNode root=new TreeNode(pre[startPre]);
          
        for(int i=startIn;i<=endIn;i++)
            if(in[i]==pre[startPre]){
                root.left=reConstructBinaryTree(pre,startPre+1,startPre+i-startIn,in,startIn,i-1);
                root.right=reConstructBinaryTree(pre,i-startIn+startPre+1,endPre,in,i+1,endIn);
            }
                  
        return root;
    }
}

c++：

/**
 * 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* constructCore(int* preStart, int* preEnd, int* vinStart, int* vinEnd){
        int rootValue = preStart[0];
        TreeNode* root = new TreeNode(preStart[0]);
        root->left = nullptr;
        root->right = nullptr;
        if(preStart == preEnd){
            if(vinStart == vinEnd && preStart == vinStart)
                return root;
        }
        int* rootVin = vinStart;
        while(rootVin <= vinEnd && *rootVin != rootValue ){
            rootVin++;
        }
        int leftLength = rootVin - vinStart;
        int* leftPreEnd = preStart + leftLength;
        if(leftLength > 0){
           root->left = constructCore(preStart + 1, leftPreEnd, vinStart, rootVin - 1);
        }
        if(leftLength < preEnd - preStart){
            root->right = constructCore(leftPreEnd + 1, preEnd, rootVin + 1, vinEnd);
            
        }
        return root;
    }
    
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
           if(pre.size() == 0 || vin.size() == 0)
               return nullptr;
           int* preStart = &pre[0];
           int* preEnd = &pre[pre.size() - 1];
           int* vinStart = &vin[0];
           int* vinEnd = &vin[vin.size() - 1];
        return constructCore(preStart, preEnd, vinStart, vinEnd);          
    }
   
    
};

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