思路:node
前序遍历的第一个节点就是树的根节点,因此咱们先根据前序遍历序列的第一个数字建立根结点,接下来在中序遍历序列中找到根结点的位置,根节点的左边就是左子树,右边就是右子树,这样就能肯定左、右子树结点的数量。在前序遍历和中序遍历的序列中划分了左、右子树结点的值以后,就能够递归地去分别构建它的左右子树。ide
/** * 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 { private: TreeNode* buildTreeCore(const vector<int>& per, int pre_start, int per_end, const vector<int>& vin, int vin_start, int vin_end) { if(pre_start > per_end || vin_start > vin_end) { return nullptr; } TreeNode *root = new TreeNode(per[pre_start]); for(int i = vin_start; i <= vin_end; ++i) { if (per[pre_start] == vin[i]) { root->left = buildTreeCore(per, pre_start + 1, pre_start + i - vin_start, vin, vin_start, i - 1); root->right = buildTreeCore(per, pre_start + i - vin_start + 1, per_end, vin, i + 1, vin_end); break; } } return root; } public: TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { if (preorder.size() != inorder.size()) { return nullptr; } return buildTreeCore(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1); } };