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Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
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思路:后续遍历的最后一个元素就是根节点,通过这个根节点,就可以把中序遍历的元素序列划分为左右子树另个部分,确定左右左子树建立的中序遍历和后续遍历元素下标范围,可以通过这个范围递归调用函数help,继续建立左右子树,最后将左右子树和root建立连接,返回root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* buildTree(vector<int>& inorder, vector<int>& postorder) {
if(inorder.size()==0)//节点数目为0,直接返回NULL
return NULL;
return help(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
}
TreeNode * help(vector<int>& inorder,int start1,int end1, vector<int>& postorder,int start2,int end2){
if(start1>end1)//下标越界,返回NULL
return NULL;
int val = postorder[end2];//取出根节点的值
TreeNode * root =new TreeNode (val);//建立根节点
int i;
for(i=start1;i<=end1;i++)
if(inorder[i]==val)
break;//此处是根节点在中序遍历的位置
int length=i-start1;//左子树元素个数
root->left =help(inorder,start1,i-1,postorder,start2,start2+length-1);//中序遍历元素左子树下标范围[start1,i-1],后续遍历左子树元素范围[start2+length-1]
root->right=help(inorder,i+1,end1,postorder,start2+length,end2-1);//中序遍历右子树下标范围[i+1,end1],后续遍历右子树元素范围[start2+length,end2-1]
return root;
}
};
106.Construct Binary Tree from Inorder and Postorder Traversal
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原文地址:http://www.cnblogs.com/zhoudayang/p/5042717.html