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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/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class
Solution {public: TreeNode *DFS(vector<int> &inorder, vector<int> &postorder,int
leftstart,int
leftend,int
rightstart,int
rightend) { if(leftend<leftstart||rightend<rightend) return
NULL; TreeNode *root=new
TreeNode(postorder[rightend]); // int
pos; for(int
i=leftstart;i<=leftend;i++) { if(inorder[i]==postorder[rightend]) { pos=i; break; } } int
len=pos-leftstart; root->left=DFS(inorder,postorder,leftstart,pos-1,rightstart,rightstart+len-1); root->right=DFS(inorder,postorder,pos+1,leftend,rightstart+len,rightend-1); return
root; } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { return
DFS(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1); }}; |
[LeetCode]Construct Binary Tree from Inorder and Postorder Traversal,布布扣,bubuko.com
[LeetCode]Construct Binary Tree from Inorder and Postorder Traversal
原文：http://www.cnblogs.com/Rosanna/p/3594878.html