LeetCode-- Construct Binary Tree from Preorder and Inorder Traversal

这是一个二叉树
               8
             /   \
           6     7
          /     /
         4     5
          \   / \
           1  2  3

Preorder    : {8} 6 4 1 7 5 2 3  
Inorder     : 4 1 6 {8} 2 5 3 7  
PostOrder   : 1 4 6 2 3 5 7 {8} 

只可能从Preorder和Inorder 或者 Postorder和Inorder的组合中构造出二叉树。

例如 Preorder 和 Inorder的组合

第一步 ： 将{8}为根进行划分
分成 {8} , {6 4 1} , {7 5 2 3}
和 {4 1 6} , {8} , {2 5 3 7}

第二步 ： 同理
将左边 {6 4 1} 和 {4 1 6}, 以{6}为根进行划分。
右边 {7 5 2 3} 和 {2 5 3 7}, 以{7}为根进行划分。

…

最终划分到单个元素集合为止。

Construct Binary Tree from Preorder and Inorder Traversal

class Solution {
public:
    TreeNode* buildTree(vector<int> & preorder, vector<int>& inorder, int pre_start, int pre_end, int in_start, int in_end){

        if(pre_end<pre_start) return NULL;

        TreeNode *node = new TreeNode(preorder.at(pre_start));

        //if(pre_start==pre_end) return node;

        int i;
        for(i=0; i<inorder.size(); i++){
            if(node->val==inorder.at(in_start+i)){
                break;
            }
        }

        node->left = buildTree(preorder, inorder, pre_start+1, pre_start+i, in_start, in_start+i-1);
        node->right = buildTree(preorder, inorder, pre_start+i+1, pre_end, in_start+i+1, in_end);

        return node;
    }


    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.size()<=0) return NULL;
        if(preorder.size()!=inorder.size()) return NULL;

        return buildTree(preorder, inorder, 0, preorder.size()-1, 0, inorder.size());
    }
};

Construct Binary Tree from Inorder and Postorder Traversal

class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size()==0) return NULL;
        if(inorder.size()!=postorder.size()) return NULL;

        return buildTree(inorder, postorder, 0, inorder.size()-1, 0, postorder.size()-1);
    }

    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder, int in_start, int in_end, int post_start, int post_end){
        if(post_end<post_start) return NULL;

        TreeNode* node = new TreeNode(postorder.at(post_end));

        //if(post_end=post_start) return node;

        int i;
        for(i=0; i<inorder.size(); i++){
            if (node->val == inorder.at(in_start+i)){
                break;
            }
        }

        node->left = buildTree(inorder, postorder, in_start, in_start+i-1, post_start, post_start+i-1);
        node->right = buildTree(inorder, postorder, in_start+i+1, in_end, post_start+i, post_end-1);

        return node;
    }
};