从树的中序遍历+前/后序遍历重建一棵树。
必须使用iterator才能过，否则会MLE。
1、preorder + inorder
第一个版本，使用坐标范围：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode *build(vector<int> &preorder, int startPre, int endPre, 13 vector<int> &inorder, int startIn, int endIn) { 14 if (startPre >= preorder.size() || endPre > preorder.size() 15 || startIn >= inorder.size() || endIn > inorder.size() 16 || startPre >= endPre || startIn >= endIn) { 17 return NULL; 18 } 19 TreeNode *root = new TreeNode(0); 20 int pos = -1; 21 for (int i = startIn; i < endIn; i++) { 22 if (inorder[i] == preorder[startPre]) { 23 pos = i - startIn; 24 break; 25 } 26 } 27 if (pos == -1) { 28 return NULL; 29 } else { 30 //vector<int> preLeft(preorder.begin() + 1, preorder.begin() + pos + 1); 31 //vector<int> preRight(preorder.begin() + pos + 1, preorder.end()); 32 //vector<int> inLeft(inorder.begin(), inorder.begin() + pos); 33 //vector<int> inRight(inorder.begin() + pos + 1, inorder.end()); 34 root->val = inorder[pos + startIn]; 35 root->left = build(preorder, startPre + 1, startPre + pos + 1, inorder, startIn, startIn + pos); 36 root->right = build(preorder, startPre + pos + 1, endPre, inorder, startIn + pos + 1, endIn); 37 } 38 return root; 39 } 40 TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { 41 return build(preorder, 0, preorder.size(), inorder, 0, inorder.size()); 42 } 43 };
第二个版本，使用迭代器：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 typedef vector<int>::iterator Iter; 13 TreeNode *build(Iter preBegin, Iter preEnd, Iter inBegin, Iter inEnd) { 14 if (preBegin == preEnd || inBegin == inEnd) { 15 return NULL; 16 } 17 TreeNode *root = new TreeNode(0); 18 Iter target = find(inBegin, inEnd, *preBegin); 19 int pos = target - inBegin; 20 if (target == inEnd || pos < 0) { 21 return NULL; 22 } else { 23 //vector<int> preLeft(preorder.begin() + 1, preorder.begin() + pos + 1); 24 //vector<int> preRight(preorder.begin() + pos + 1, preorder.end()); 25 //vector<int> inLeft(inorder.begin(), inorder.begin() + pos); 26 //vector<int> inRight(inorder.begin() + pos + 1, inorder.end()); 27 root->val = (*preBegin); 28 root->left = build(preBegin + 1, preBegin + pos + 1, inBegin, inBegin + pos); 29 root->right = build(preBegin + pos + 1, preEnd, inBegin + pos + 1, inEnd); 30 } 31 return root; 32 } 33 TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { 34 return build(preorder.begin(), preorder.end(), inorder.begin(), inorder.end()); 35 } 36 };
CE了无数次，RE了好多次是由于边界的问题。
WA是由于使用坐标边界的计算错误。。。
总之就是一个经典题，有很多细节值得深入研究。
2、postorder + inorder
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 typedef vector<int>::iterator Iter; 13 14 TreeNode *build(Iter inBegin, Iter inEnd, Iter postBegin, Iter postEnd) { 15 if (inBegin == inEnd || postBegin == postEnd) { 16 return NULL; 17 } 18 Iter target = find(inBegin, inEnd, *(postEnd-1)); 19 int dis = target - inBegin; 20 if (target == inEnd || dis < 0) { 21 return NULL; 22 } 23 TreeNode * root = new TreeNode(*target); 24 root->left = build(inBegin, inBegin + dis, postBegin, postBegin + dis); 25 root->right = build(inBegin + dis + 1, inEnd, postBegin + dis, postEnd - 1); 26 return root; 27 } 28 TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { 29 return build(inorder.begin(), inorder.end(), postorder.begin(), postorder.end()); 30 } 31 };
CE了一次，少写了一个分号，不开心。。。
用迭代器写的思路还是非常清晰的。
[Leetcode][Tree][Construct Binary Tree from Preorder/Postorder and Inorder Traversal ]
原文地址：http://www.cnblogs.com/poemqiong/p/3815330.html
