LeetCode106 Construct Binary Tree from Inorder and Postorder Traversal

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

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

 

分析：

跟前一题方法一样，只需要分析出中序与后序恢复时，需从后序的最后一个元素读出来当做根节点，在中序中寻找其位置。

其余递归过程跟前序中序恢复一致。

代码：

/**
 * 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 {
private:
    TreeNode* helper(vector<int>& inorder, int inStart, int inEnd, vector<int>& postorder, int postStart, int postEnd) {
        int sz = inEnd - inStart;
        if (sz == 0) {
            return nullptr;
        }
        int rootVal = postorder[postEnd - 1];
        TreeNode* root = new TreeNode(rootVal);
        int pos;
        for (pos = 0; pos < sz; ++pos) {
            if (inorder[inStart + pos] == rootVal) {
                break;
            }
        }
        if (pos < 0) {
            root -> left = nullptr;
        }
        else {
            root -> left = helper(inorder, inStart , inStart + pos, postorder, postStart, postStart + pos);            
        }
        if (pos + 1 >= sz) {
            root -> right = nullptr;
        }
        else {
            root -> right = helper(inorder, inStart + pos + 1, inEnd, postorder, postStart + pos, postEnd - 1);  
        }
        return root;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return helper(inorder, 0, inorder.size(), postorder, 0, postorder.size());
    }
};