leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal

1.题目

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree

2.思路

递归.每次从后续遍历最后得到root，然后根据前序分割left tree和right tree,之后递归.

3.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 {
// inorder: left-root-right
//postorder: left-right-root
public:
TreeNode* create(vector<int> &inorder, vector<int> &postorder, int is, int ie, int ps, int pe){
if(ps > pe){
return nullptr;
}
TreeNode* node = new TreeNode(postorder[pe]);
//找到inorder中root位置
int pos;
for(int i = is; i <= ie; i++){
if(inorder[i] == node->val){
pos = i;
break;
}
}
node->left = create(inorder, postorder, is, pos - 1, ps, ps + pos - is - 1);
node->right = create(inorder, postorder, pos + 1, ie, pe - ie + pos, pe - 1);
return node;
}

TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
return create(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1);
}
};