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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
题目大意：
由二叉树的前序和中序遍历得出二叉树的结构。
解题思路：
对preorder进行依次遍历，确定当前的节点；对inorder需要找出当前节点的位置，将其分为两部分；
/**
* 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:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return buildTreehelper(preorder, 0, preorder.size(), inorder, 0, inorder.size());
}
TreeNode* buildTreehelper(vector<int>& preorder, int sp, int ep, vector<int>& inorder, int si, int ei){
if(sp == ep) return nullptr;
TreeNode* tmp = new TreeNode(preorder[sp]);
int idx = find(inorder.begin()+si, inorder.begin()+ei, preorder[sp]) - inorder.begin() - si;
// 此处依然遵循左开右闭原则
tmp->left = buildTreehelper(preorder, sp+1, sp+1+idx, inorder, si, si+idx);
tmp->right = buildTreehelper(preorder, sp+1+idx, ep, inorder, si+idx+1, ei);
return tmp;
}
};