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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.
For example, given
inorder = [9,3,15,20,7]postorder = [9,15,7,20,3]
Return the following binary tree:
3 / \ 9 20 / \ 15 7
题意:根据一棵树的中序遍历与后序遍历构造二叉树。可以假设树中没有重复的元素。
简单的分治题目,递归解决。代码如下:
/** * 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 & inorder, vector & postorder) { return rebuild(0, inorder.size() - 1, 0, postorder.size() - 1, inorder, postorder); } TreeNode* rebuild(int leftin, int rightin, int leftpost, int rightpost, vector & inorder, vector & postorder) { if (leftin > rightin || leftpost > rightpost) return nullptr; TreeNode *root = new TreeNode(postorder[rightpost]); int rootin = leftin; while (rootin <= rightin && inorder[rootin] != postorder[rightpost]) ++rootin; int leftsize = rootin - leftin; root->left = rebuild(leftin, rootin - 1, leftpost, leftpost + leftsize - 1, inorder, postorder); root->right = rebuild(rootin + 1, rightin, leftpost + leftsize, rightpost - 1, inorder, postorder); return root; }};
效率如下:
执行用时:52 ms, 在所有 C++ 提交中击败了26.98% 的用户内存消耗:16.9 MB, 在所有 C++ 提交中击败了85.71% 的用户
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