LintCode-72.中序遍历和后序遍历树构造二叉树

中序遍历和后序遍历树构造二叉树

根据中序遍历和后序遍历树构造二叉树

注意事项

你可以假设树中不存在相同数值的节点

样例

给出树的中序遍历： [1,2,3] 和后序遍历： [1,3,2]
返回如下的树：
  2
 /  
1   3

标签

二叉树

code

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
    /**
     *@param inorder : A list of integers that inorder traversal of a tree
     *@param postorder : A list of integers that postorder traversal of a tree
     *@return : Root of a tree
     */
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        // write your code here
        TreeNode *root = NULL;
        vector<int> inorder_l,inorder_r,postorder_l,postorder_r;
        int i,root_index=0;
        int size = postorder.size();

        if(inorder.empty()!=1 || postorder.empty()!=1) {
            root = new TreeNode(postorder[size-1]); //  在后序队列中找根节点

            //  在中序队列中找出根节点位置
            for(i=0; i<inorder.size(); i++) {
                if(postorder[size-1] == inorder[i])
                    break;
                root_index++;
            }

            //  左右子树的前序、中序队列
            for(i=0; i<root_index; i++) {
                postorder_l.push_back(postorder[i]);
                inorder_l.push_back(inorder[i]);
            }
            for(i=root_index+1; i<inorder.size(); i++) {
                postorder_r.push_back(postorder[i-1]);
                inorder_r.push_back(inorder[i]);
            }

            root->left = buildTree(inorder_l, postorder_l);
            root->right = buildTree(inorder_r, postorder_r);
        }
        return root;
    }
};