剑指 Offer 07. 重建二叉树

题目链接

前序遍历：根节点->左子树->右子树
中序遍历：左子树->根节点->右子树
前序遍历的第一个节点是根节点，然后根据根节点在中序遍历数组中的位置，该位置左边是左子树的节点数量，该位置右边是右子树的节点数量。
C++:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    unordered_map<int, int> map;
    vector<int> preorder;
    TreeNode* DFS(int prestart, int preend, int instart, int inend)
    {
        if(prestart > preend)
            return nullptr;
        int num = preorder[prestart];
        TreeNode* root = new TreeNode(num);
        if(prestart == preend) //如果该树只有一个节点
            return root;
        int rootindex = map[num];
        int leftcount = rootindex-instart;
        int rightcount = inend-rootindex;
        TreeNode* leftnode = DFS(prestart+1, prestart+leftcount, instart, rootindex-1);
        TreeNode* rightnode = DFS(preend-rightcount+1, preend, rootindex+1, inend);
        root->left = leftnode;
        root->right = rightnode;
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int prelen = preorder.size();
         if(prelen == 0) {
            return nullptr;
        }
        this->preorder = preorder;
        for(int i = 0; i < prelen; i++)
        {
            map[inorder[i]] = i;
        }
        return DFS(0, prelen-1, 0, prelen-1);
    }

};

官方给的java代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
        int length = preorder.length;
        for (int i = 0; i < length; i++) {
            indexMap.put(inorder[i], i);
        }
        TreeNode root = buildTree(preorder, 0, length - 1, inorder, 0, length - 1, indexMap);
        return root;
    }

    public TreeNode buildTree(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd, Map<Integer, Integer> indexMap) {
        if (preorderStart > preorderEnd) {
            return null;
        }
        int rootVal = preorder[preorderStart];
        TreeNode root = new TreeNode(rootVal);
        if (preorderStart == preorderEnd) {
            return root;
        } else {
            int rootIndex = indexMap.get(rootVal);
            int leftNodes = rootIndex - inorderStart, rightNodes = inorderEnd - rootIndex;
            TreeNode leftSubtree = buildTree(preorder, preorderStart + 1, preorderStart + leftNodes, inorder, inorderStart, rootIndex - 1, indexMap);
            TreeNode rightSubtree = buildTree(preorder, preorderEnd - rightNodes + 1, preorderEnd, inorder, rootIndex + 1, inorderEnd, indexMap);
            root.left = leftSubtree;
            root.right = rightSubtree;
            return root;
        }
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/solution/mian-shi-ti-07-zhong-jian-er-cha-shu-by-leetcode-s/
来源：力扣（LeetCode）
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不递归不知道C++传vector是多么浪费时间！！！
这里有一个坑，就是如果C++版本像java一样传递数组preorder， inorder，还有hashmap，会在最后一个数据超时。

如果不传hashmap:

如果也不传inorder：

如果也不传preorder：

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