标签:
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路:
数据结构的基本题目,根据中序遍历和后序遍历确定整个树的结构。我的方法是通过后续遍历确定根节点是什么,在中序遍历中寻找这个节点的位置,那么它的左边就是左子树,右边就是右子树,如此递归寻找。这里的一个难点是判断下一次递归中左子树的根节点,本代码中就是index-r+i-1,这个取值是我试出来的,我也不知道为什么。
题解:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void build(vector<int> &inorder, vector<int> &postorder, int l, int r, int index, TreeNode *&root) { if(l>r) return; root = new TreeNode(postorder[index]); int i; for(i=l;i<=r;i++) if(inorder[i]==postorder[index]) break; build(inorder, postorder, i+1, r, index-1, root->right); build(inorder, postorder, l, i-1, index-r+i-1, root->left); } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { if(inorder.size()==0) return NULL; TreeNode *root; build(inorder, postorder, 0, inorder.size()-1, postorder.size()-1, root); return root; } };
[leetcode] Construct Binary Tree from Inorder and Postorder Traversal
标签:
原文地址:http://www.cnblogs.com/jiasaidongqi/p/4213417.html