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, 0, inorder.size() - 1);
    }
    
    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 - 1);
        root->right = recursive(preorder, inorder, i + 1, 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, 0, inorder.size() - 1);
    }
    
    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 + 1, e);
        root->left = recursive(postorder, inorder, s, i - 1);
    }
};