[LeetCode][105,106] Construct Binary Tree from Inorder and (Post/Pre)order Traversal

Description:

Given (pre/post)order and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

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Solution:

题意：通过前序遍历和中序遍历，或后序遍历和中序遍历，构造一棵二叉树。

思路：递归构造。

由前序遍历和中序遍历：

/** * 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<int>& preorder, vector<int>& inorder) {        this->preorder = preorder;        this->inorder = inorder;        return constructTree(0, 0, inorder.size() - 1);    }    TreeNode* constructTree(int preStart, int inStart, int inEnd) {        if (preStart >= preorder.size() || inStart > inEnd)            return NULL;        int curIndex = 0;        TreeNode* curRoot = new TreeNode(preorder[preStart]);        for (int i = inStart; i <= inEnd; i++) {            if (preorder[preStart] == inorder[i]) {                curIndex = i;                break;            }        }        curRoot->left = constructTree(preStart + 1, inStart, curIndex - 1);        curRoot->right = constructTree(preStart + curIndex - inStart + 1, curIndex + 1, inEnd);        return curRoot;    }private:    vector<int> preorder, inorder;};

由后序遍历和中序遍历：

/** * 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<int>& inorder, vector<int>& postorder) {        this->postorder = postorder;        this->inorder = inorder;        return constructTree(postorder.size() - 1, 0, inorder.size() - 1);    }    TreeNode* constructTree(int postStart, int inStart, int inEnd) {        if (postStart < 0 || inStart > inEnd)            return NULL;        int curIndex = 0;        TreeNode* curRoot = new TreeNode(postorder[postStart]);        for (curIndex = inStart; curIndex <= inEnd; curIndex++)             if (postorder[postStart] == inorder[curIndex])                 break;                curRoot->left = constructTree(postStart - (inEnd - curIndex) - 1, inStart, curIndex - 1);        curRoot->right = constructTree(postStart - 1, curIndex + 1, inEnd);        return curRoot;    }private:    vector<int> postorder, inorder;};