題目描述

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

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

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

    void ConstructTree(TreeNode *&tree, vector<int> &inorder, int inbegin, int inend, vector<int> &postorder, int postbegin, int postend)
    {
        if(inbegin > inend || postbegin > postend)
            return;

        int i;
        int mid = postorder[postend];
        tree = new TreeNode(mid);
        for(i = inbegin; i <= inend; i++)
        {
            if(inorder[i] == mid)
                break;
        }
        ConstructTree(tree->left, inorder, inbegin, i-1, postorder, postbegin, postbegin+i-1-inbegin);
        ConstructTree(tree->right, inorder, i+1, inend, postorder, postbegin+i-inbegin, postend-1);


    }


    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
    {
        if(inorder.size() == 0 || postorder.size() == 0)
            return NULL;
        TreeNode *tree = NULL;
        ConstructTree(tree, inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);

        return tree;
    }
};

[leetcode] construct-binary-tree-from-inorder-and-postorder-