(二叉树 递归) leetcode 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 -------------------------------------------------------------------------------------
就是从中序遍历和后序遍历构建二叉树。可以用递归方式。注意递归的终止条件。
和leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal几乎一样。 参考博客:http://www.cnblogs.com/grandyang/p/4296193.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>& inorder, vector<int>& postorder) {
return build(inorder,,inorder.size()-,postorder,,postorder.size() - );
}
TreeNode *build(vector<int> &inorder,int ileft,int iright,vector<int> &postorder,int pleft,int pright){
if(ileft > iright ||pleft > pright) return NULL; //终止条件,就是当序列的长度为0时,递归终止。
int i = ;
TreeNode *cur = new TreeNode(postorder[pright]);
for(i = ileft; i < inorder.size(); i++){
if(inorder[i] == cur->val)
break;
}
cur->left = build(inorder,ileft,i-,postorder,pleft,pleft + i - ileft - );
cur->right = build(inorder,i + ,iright,postorder,pleft + i - ileft,pright - );
return cur;
}
};
还有一个方法,就是建立几个数组,保存分割后的数组。不过时间会很长。
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>& inorder, vector<int>& postorder) {
return build(inorder,postorder);
}
TreeNode *build(vector<int> &inorder,vector<int> &postorder){
if(inorder.size() == || postorder.size() == ) //递归条件,当然也可以加上if(inorder.size() == 1 || postorder.size() == 1)return cur;这个递归条件。
return NULL;
int rootval = postorder.back();
TreeNode *cur = new TreeNode(rootval);
int i = ;
for(i = ; i < inorder.size(); i++){
if(inorder[i] == rootval) break;
}
vector<int> inleft,inright;
vector<int> poleft,poright;
for(int j = ; j < i; j++){
inleft.push_back(inorder[j]);
poleft.push_back(postorder[j]);
}
for(int j = i + ; j < inorder.size(); j++){
inright.push_back(inorder[j]);
}
for(int j = i; j < postorder.size() - ; j++){
poright.push_back(postorder[j]);
}
cur->left = build(inleft,poleft);
cur->right = build(inright,poright);
return cur;
}
};
这两个方法从算法上看是一样的,只是代码的实现不同而已。
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