Leetcode | Construct Binary Tree from Inorder and (Preorder or Postorder) Traversal

Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

递归构建。

思路就是： preorder可以定位到根结点，inorder可以定位左右子树的取值范围。

1. 由preorder得到根结点；把preorder第一个点删掉；

2. 先建左子树；再建右子树；

通过一个区间来表示左右子树的取值范围。因为inorder左右子树的范围都是连续的。中间就是root。

 class Solution {
 public:
     TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
         return recursive(preorder, inorder, , inorder.size() - );
     }
 
     TreeNode* recursive(vector<int> &preorder, vector<int> &inorder, int s, int e) {
         if (s > e) return NULL;
         if (preorder.empty()) return NULL;
         TreeNode *root = new TreeNode(preorder.front());
         preorder.erase(preorder.begin());
 
         int i = s;
         for (; i <= e && inorder[i] != root->val; ++i);
         root->left = recursive(preorder, inorder, s, i - );
         root->right = recursive(preorder, inorder, i + , e);
     }
 };

Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

和上面类似。有两点不同。

1. postorder，最后一个元素是根结点。

2. 先构建右子树，再构建左子树。

 class Solution {
 public:
     TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
         return recursive(postorder, inorder, , inorder.size() - );
     }
 
     TreeNode* recursive(vector<int> &postorder, vector<int> &inorder, int s, int e) {
         if (s > e) return NULL;
         if (postorder.empty()) return NULL;
         TreeNode *root = new TreeNode(postorder.back());
         postorder.pop_back();
 
         int i = s;
         for (; i <= e && inorder[i] != root->val; ++i);
         root->right = recursive(postorder, inorder, i + , e);
         root->left = recursive(postorder, inorder, s, i - );
     }
 };