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