剑指offer-4-重建二叉树

川长思鸟来 2022-02-18 20:26 160阅读 0赞

**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。

【解题思路】  
（1）前序遍历序列中的首元素为根节点  
（2）中序遍历序列中根节点元素之前的为左子树，之后的为右子树  
（3）对左子树和右子树进行递归直到前序和中序序列的长度都为1

/**
     * 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 rootValue = pre[0];
    		TreeNode * root = new TreeNode(rootValue);
    
    		if (pre.size() == 1)
    		{
    			if (vin.size() == 1 && pre[0] == vin[0])
    			{
    				return root;
    			}
    		}
    
    		int rootVin = 0;
    		for (int i = 0; i < vin.size();++i)
    		{
    			if (root->val == vin[i])
    			{
    				rootVin = i;
    				break;
    			}
    		}
    
    
    		if (rootVin > 0)//left tree
    		{
    			vector<int> preLeft,vinLeft;
    			for (int i = 0; i < rootVin; i++)
    			{
    				preLeft.push_back(pre[i+1]);
    				vinLeft.push_back(vin[i]);
    			}
    			root->left = reConstructBinaryTree(preLeft, vinLeft);
    		}
    		if(rootVin < vin.size()-1)
    		{
    			vector<int> preRight, vinRight;
    			for (int i = rootVin+1; i < vin.size(); i++)
    			{
    				preRight.push_back(pre[i]);
    				vinRight.push_back(vin[i]);
    			}
    			root->right = reConstructBinaryTree(preRight, vinRight);
    		}
    
    		return root;
        }
    };