题解 | #重建二叉树#

• 前序和中序确定二叉树
• 前序数组的第一项为根节点；
• 根据前序数组的第一项确定中序数组的分割点，分为左中序数组和右中序数组；
• 根据左中序数组的大小，将前序数组分为左前序数组和右中序数组；
• 左节点由左前序数组和左中序数组确定；
• 右节点由右前序数组和右中序数组确定。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    TreeNode* traversal(vector<int>& pre, vector<int>& vin) {
        if (pre.size() == 0) return NULL;
        int rootValue = pre[0];
        TreeNode* root = new TreeNode(rootValue);
        if (pre.size() == 1) return root;
        int splitIndex;
        for (splitIndex = 0; splitIndex < vin.size(); splitIndex++) {
            if (vin[splitIndex] == rootValue) break;
        }
        vector<int> leftVin(vin.begin(), vin.begin() + splitIndex);
        vector<int> rightVin(vin.begin() + splitIndex + 1, vin.end());
        vector<int> leftPre(pre.begin() + 1, pre.begin() + leftVin.size() + 1);
        vector<int> rightPre(pre.begin() + leftVin.size() + 1, pre.end());
        
        root->left = traversal(leftPre, leftVin);
        root->right = traversal(rightPre, rightVin);
        return root;
    }
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        if (pre.size() == 0 || vin.size() == 0) return NULL;
        return traversal(pre, vin);

    }
};