一、
题目：
105. 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.
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
解答：
/**
* 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) {
if(preorder.size() != inorder.size() || preorder.empty() || inorder.empty() )
return NULL;
TreeNode * inRoot = myHelper(preorder, inorder, 0, 0, inorder.size() - 1);
return inRoot;
}
//递归的调用myHelper（），先构建根的左子树，再构建根的右子树
TreeNode* myHelper(vector<int> & preorder, vector<int>& inorder, int preRootPos, int inLeftPos, int inRightPos)//前序的根节点位置preRootPos,此根节点在中序中的左边界位置inLeftPos, 右边界位置inRightPos
{
if(inLeftPos > inRightPos)
return NULL;
int inRootPos, rootVal;
rootVal = preorder[preRootPos];
//寻找根节点rootVal在inorder中的位置inRootPos
for(int i = inLeftPos; i <= inRightPos; ++i)//int inLeftPos, int inRightPos 都是索引 <=
{
if(rootVal == inorder[i])
inRootPos = i;
}
//计算inRootVal到inLeftPOS的长度，即rootVal为根的左子树的长度
int leftLen = inRootPos - inLeftPos;
TreeNode* inRoot = new TreeNode(rootVal);
inRoot->left = myHelper(preorder, inorder, preRootPos + 1, inLeftPos, inRootPos - 1);
inRoot->right = myHelper(preorder, inorder, preRootPos + leftLen + 1, inRootPos + 1, inRightPos);
return inRoot;
}
};二、
题目：
106. 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.
For example, given
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
解答：
/**
* 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>& inorder, vector<int>& postorder) {
if(inorder.size() != postorder.size() || inorder.empty() || postorder.empty())
return nullptr;
TreeNode * inRoot;
inRoot = myHelper(inorder, postorder, postorder.size() - 1, 0, inorder.size() - 1);
return inRoot;
}
TreeNode* myHelper(vector<int>& inorder, vector<int> & postorder, int postRootPos, int inLeftPos, int inRightPos)
{
if(inLeftPos > inRightPos)
return nullptr;
int inRootPos, rootVal;
rootVal = postorder[postRootPos];
for(int i = inLeftPos; i <= inRightPos; ++i)
if(rootVal == inorder[i])
inRootPos = i;
int rightLen = inRightPos - inRootPos;
TreeNode* inRoot = new TreeNode(rootVal);
inRoot->left = myHelper(inorder, postorder, postRootPos - rightLen - 1, inLeftPos, inRootPos - 1);
inRoot->right = myHelper(inorder, postorder, postRootPos - 1, inRootPos + 1, inRightPos);
return inRoot;
}
};