LeetCode-Construct Binary Tree from Inorder and Postorder Traversal

数据结构的基础知识，根据中序、后序遍历或者先序、中序遍历构建二叉树；

典型的递归；

中序、后序的代码：

/**
 * Definition for binary tree
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = inorder.size();
        int n = postorder.size();
        if (m != n || !m || !n) {
            return NULL;
        }
        return build(inorder, 0, m - 1, postorder, 0, n - 1);
    }
    TreeNode *build(vector<int> &inorder, int m1, int m2, vector<int> &postorder, int n1, int n2) {
        if (m1 > m2) {
            return NULL;
        }
        TreeNode* node = new TreeNode(postorder[n2]);
        int i;
        for (i = m1; i <= m2; ++i) {
            if (inorder[i] == postorder[n2]) {
                break;
            }
        }
        node->left = build(inorder, m1, i - 1, postorder, n1, n1 + i - 1 - m1);
        node->right = build(inorder, i + 1, m2, postorder, n2 - m2 + i, n2 - 1);
        return node;
    }
};

先序、中序的代码：

/**
 * Definition for binary tree
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = preorder.size();
        int n = inorder.size();
        if (m != n || !m || !n) {
            return NULL;
        }
        return build(preorder, 0, m - 1, inorder, 0, n - 1);
    }
    TreeNode *build(vector<int> &preorder, int m1, int m2, vector<int> &inorder, int n1, int n2) {
        if (m1 > m2) {
            return NULL;
        }
        int i;
        for(i = n1; i <= n2; ++i) {
            if (inorder[i] == preorder[m1]) {
                break;
            }
        }
        TreeNode* node = new TreeNode(preorder[m1]);
        node->left = build(preorder, m1 + 1, i - n1 + m1, inorder, n1, i - 1);
        node->right = build(preorder, i - n1 + m1 + 1, m2, inorder, i + 1, n2);
        return node;
    }
};