https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/
思路:先序遍历:根左右。中序遍历:左根右。那么可以由先序遍历得到根节点的值,依据此值再找到根节点在中序遍历的位置,那么就可以分出左子树和右子树,分治下去即可。
/**
* 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* dfs(int l1,int r1,int l2,int r2,vector<int>& preorder,vector<int>& inorder)
{
if(l1==r1)//只有一个节点时
{
TreeNode *node=new TreeNode(preorder[l1]);
return node;
}
int rootval=preorder[l1];//根节点的值
int i=l2;
while(i<=r2&&inorder[i]!=rootval)//在中序遍历中找到根节点
++i;
int lnum=i-l2;//左子树元素个数
int rnum=r2-i;//右子树元素个数
TreeNode *root=new TreeNode(rootval);
TreeNode *lc=NULL,*rc=NULL;
if(lnum)
lc=dfs(l1+1,l1+lnum,l2,i-1,preorder,inorder);
if(rnum)
rc=dfs(r1-rnum+1,r1,i+1,r2,preorder,inorder);
root->left=lc;
root->right=rc;
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int siz=preorder.size();
if(!siz)
return NULL;
return dfs(0,siz-1,0,siz-1,preorder,inorder);
}
};
非递归版?感觉这个有点难理解,最好结合例子体会一下。
/**
* 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) {
int siz=preorder.size();
if(!siz)
return NULL;
stack<TreeNode*> s;
int j=0;
TreeNode *node,*head=new TreeNode(preorder[0]);
s.push(head);
for(int i=1;i<siz;i++)
{
if(s.top()->val!=inorder[j])
{
node=new TreeNode(preorder[i]);
s.top()->left=node;
s.push(node);
}
else
{
while(!s.empty()&&s.top()->val==inorder[j])
node=s.top(),s.pop(),j++;
TreeNode *tmp=new TreeNode(preorder[i]);
node->right=tmp;
s.push(tmp);
}
}
return head;
}
};