(二叉树 递归) leetcode 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
-----------------------------------------------------------------------------------
就是从先序遍历和中序遍历构建踹个二叉树,可以用递归方式,注意递归的终止条件。
和leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal几乎一样。
参考博客:http://www.cnblogs.com/grandyang/p/4296500.html
C++代码:
/**
* 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) {
return build(preorder,,preorder.size() - ,inorder,,inorder.size() - );
}
TreeNode* build(vector<int> &preorder,int pleft,int pright,vector<int> &inorder,int ileft,int iright){
if(pleft > pright || ileft > iright) return NULL; //递归的终止条件。
int i = ;
TreeNode *cur = new TreeNode(preorder[pleft]);
for(i = ileft; i < inorder.size(); i++){
if(inorder[i] == cur->val)
break;
}
cur->left = build(preorder,pleft + ,pleft + i - ileft,inorder,ileft,i-);
cur->right = build(preorder,pleft + i - ileft + ,pright,inorder,i+,iright);
return cur;
}
};
也有一个方法:
/**
* 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) {
return build(preorder,inorder);
}
TreeNode* build(vector<int> &preorder,vector<int> &inorder){
if(preorder.size() == || inorder.size() == ) return NULL; //递归条件。当然,也还可以加上if(preorder.size() == 1 || inorder.size() == 1) return cur;这个递归条件。
int rootval = preorder[];
int i = ;
for(i = ; i < inorder.size(); i++){
if(inorder[i] == rootval)
break;
}
vector<int> inleft,inright,preleft,preright;
for(int k = ; k < i + ; k++)
preleft.push_back(preorder[k]);
for(int k = i + ; k < preorder.size(); k++){
preright.push_back(preorder[k]);
inright.push_back(inorder[k]);
}
for(int k = ; k < i; k++)
inleft.push_back(inorder[k]);
TreeNode *cur = new TreeNode(rootval);
cur->left = build(preleft,inleft);
cur->right = build(preright,inright);
return cur;
}
};
这两个方法从思想上是一样的,只不过代码的实现有所不同。
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