剑指offer_重构二叉树

Posted 2021-09-09 ~千里之行，始于足下~

重建二叉树

/**
 * 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* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        int n = pre.size();
        int m = vin.size();
        if(n!=m || n == 0)
            return NULL;
        return construct(pre, vin, 0, n-1, 0, m-1);
    }

    TreeNode* construct(vector<int>& pre, vector<int>& vin, int l1, int r1, int l2, int r2)
    {
        TreeNode* root = new TreeNode(pre[l1]);
        if(r1 == l1)
        {
            return root;
        }
        int val = pre[l1];
        int index;
        for(index = l2; index <= r2; index ++)
        {
            if(vin[index] == val)
                break;
        }
        int left_tree_len  = index - l2;
        int right_tree_len = r2 - index;
        if(left_tree_len > 0)
            root->left = construct(pre, vin, l1+1, l1+left_tree_len, l2, index-1);
        if(right_tree_len >0 )
            root->right = construct(pre, vin, l1+1+left_tree_len, r1, index+1, r2);
        return root;
    }
};

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.Arrays;
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] vin) {
        if (0 == pre.length)
            return null;
        TreeNode node = new TreeNode(pre[0]);
        int rootIndex = -1;
        for (int i = 0; i < vin.length; i++)
        {
            if(pre[0] == vin[i])
            {
                rootIndex = i;
                break;
            }
        }
        node.left = reConstructBinaryTree(Arrays.copyOfRange(pre,1,rootIndex+1),Arrays.copyOfRange(vin,0,rootIndex));
        node.right = reConstructBinaryTree(Arrays.copyOfRange(pre,rootIndex+1,pre.length),Arrays.copyOfRange(vin,rootIndex+1,vin.length));
        return node;
    }
}

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回构造的TreeNode根节点
    def reConstructBinaryTree(self, pre, vin):
        # write code here
        if not pre:
            return None;
        
        root = TreeNode(pre[0])
        tmp = vin.index(pre[0])
        root.left = self.reConstructBinaryTree(pre[1:tmp+1], vin[0:tmp])
        root.right = self.reConstructBinaryTree(pre[tmp+1:], vin[tmp+1:])
        return root