根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
代码
/**
* 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 *buildTree(vector<int> &preorder, vector<int> &inorder) {
return buildTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
}
TreeNode *buildTree(vector<int> &preorder, int pLeft, int pRight, vector<int> &inorder, int iLeft, int iRight) {
if (pLeft > pRight || iLeft > iRight) return NULL;
int i = 0;
for (i = iLeft; i <= iRight; ++i) {
if (preorder[pLeft] == inorder[i]) break;
}
TreeNode *cur = new TreeNode(preorder[pLeft]);
cur->left = buildTree(preorder, pLeft + 1, pLeft + i - iLeft, inorder, iLeft, i - 1);
cur->right = buildTree(preorder, pLeft + i - iLeft + 1, pRight, inorder, i + 1, iRight);
return cur;
}
};
