LeetCode题解： Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

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

思路：

已知中序遍历和后序遍历的结果，要求反推树结构。首先注意到后序遍历中，最后一个元素必然是树的根结点。因为树的值没有重复，所以可以在中序遍历中寻找对应的值，这个值左右部分分别是根结点的左子树和右子树，从而知道左右子树元素的个数和中序遍历的结果。再反过来，因为后序遍历首先遍历左子树，再遍历右子树，所以同样可以得到后序遍历的结果。递归构造即可得到原始树结构。

题解：

/**
 * 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(const vector<int>& inorder, int is, int ie,
                        const vector<int>& postorder, int ps, int pe)
    {
        if (is > ie)
            return nullptr;
        
        int base = postorder[pe];
        TreeNode* node = new TreeNode(base);
        
        int ipos = is;
        while(inorder[ipos] != base) ++ipos;
        
        int left_nodes = ipos - is;
        int right_nodes = ie - ipos;
        
        if (left_nodes != 0)
            node->left = buildTree(inorder, is, is + left_nodes - 1,
                                   postorder, ps, ps + left_nodes - 1);
        
        if (right_nodes != 0)
            node->right = buildTree(inorder, ipos + 1, ie,
                                   postorder, ps + left_nodes, pe - 1);
                               
        return node;
    }

    TreeNode *buildTree(const vector<int> &inorder, const vector<int> &postorder) {
        if (inorder.empty())
            return nullptr;
            
        return buildTree(inorder, 0, inorder.size() - 1,
                         postorder, 0, postorder.size() - 1);
    }
};