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珠海网站建设小程序,即时热榜,网站举报入口,网站策划专员Problem: 889. Construct Binary Tree from Preorder and Postorder Traversal 根据前序和后序遍历构造二叉树 前序遍历是【根左右】#xff0c;后序遍历是【左右根】#xff0c;所以preorder第一个一定是根节点#xff0c;postorder最后一个一定是根节点#xff0c;两者一…Problem: 889. Construct Binary Tree from Preorder and Postorder Traversal 根据前序和后序遍历构造二叉树前序遍历是【根左右】后序遍历是【左右根】所以preorder第一个一定是根节点postorder最后一个一定是根节点两者一定相等postorder倒数第二个一定是右子树的根节点所以可以根据postorder倒数第二个将前序遍历划分开来划分成左右子树前序遍历确定好右子树节点个数以后就可以将后序遍历划分开Code/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: int preL, postL; TreeNode* construct(int preLeft, int preRight, int postLeft, int postRight, vectorint preorder, vectorint postorder) { if(preLeft preRight || postLeft postRight) return nullptr; TreeNode* root new TreeNode; root-val preorder[preLeft]; if(preLeftpreRight || postLeft postRight) return root; int k preLeft 1; while(k preL preorder[k]!postorder[postRight-1]) k; root-left construct(preLeft1, k-1, postLeft, postRight - (preRight - k 1)-1, preorder, postorder); root-right construct(k, preRight, postRight - (preRight - k 1), postRight-1, preorder, postorder); return root; } TreeNode* constructFromPrePost(vectorint preorder, vectorint postorder) { TreeNode* root nullptr; preL preorder.size()-1; postL postorder.size()-1; root construct(0, preL, 0, postL, preorder, postorder); return root; } };