Topic description
Input the result of preorder traversal and inorder traversal of a binary tree , please rebuild the binary tree.
Notice:
The value of each node in the binary tree is different from each other;
the input pre-order traversal and in-order traversal must be legal;
Data range
The number of nodes in the tree ranges from [0,100].
Example
Given:
Preorder traversal is: [3, 9, 20, 15, 7]
Inorder traversal is: [9, 3, 15, 20, 7]
Returns: [3, 9, 20, null, null, 15, 7, null, null, null, null]
The returned binary tree looks like this:
3
/ \
9 20
/ \
15 7
CPP
/**
* 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:
unordered_map<int, int> hash;
TreeNode* dfs(vector<int>& pre, vector<int>& in, int pl, int pr, int il, int ir) {
if (pl > pr) return NULL;
// 用中序遍历确定,根结点左子树的大小
int k = hash[pre[pl]] - il;
TreeNode* root = new TreeNode(pre[pl]);
// 边界可以举一个实际的例子来确定
root->left = dfs(pre, in, pl + 1, pl + k, il, il + k - 1);
root->right = dfs(pre, in, pl + k + 1, pr, il + k + 1, ir);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int n = inorder.size();
for (int i = 0; i < n; i++) hash[inorder[i]] = i;
return dfs(preorder, inorder, 0, n-1, 0, n-1);
}
};