Given preorder and inorder traversal of a tree, construct the binary tree.
由树的前序遍历和中序遍历来构建子树。
首先找到根节点,然后递归找到左侧和右侧根节点完成。
/**
* 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) {
int num = preorder.size();
if(num==0)
return NULL;
return Recursion(0, num-1, 0,preorder, inorder);
}
TreeNode* Recursion(int Istart,int Iend,int Pstart,vector<int>& preorder, vector<int>& inorder){
if(Istart>Iend)
return NULL;
TreeNode* root = new TreeNode(preorder[Pstart]); //标记根的位置,特别注意根位置的变化
int mid = 0;
for(int i=Istart; i<=Iend; ++i){
if(inorder[i]==root->val)
mid = i;
}
root->left = Recursion(Istart,mid-1,Pstart+1,preorder,inorder); //递归指向左右的根
root->right = Recursion(mid+1,Iend,Pstart+mid-Istart+1,preorder,inorder);
return root;
}
};