[C++]LeetCode: 38 Construct Binary Tree from Inorder and Postorder Traversal

题目 ：

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

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

Anwser 1: 递归法

思路解析： 可以参考《Construct Binary Tree from Preorder and Inorder Traversal》

Attention:

1. 定义了全局变量后再定义局部变量将导致致命错误，与设计思路不符，并且会导致覆盖。

Error Code: 有时候是顺手敲上去的，一定要避免这种情况，代码每句一定要想好再写。

int pos;

for( int pos = sMid; pos <= eMid; pos++)
        {
            if(inorder[pos] == poval)
                break;
        } 

2. 用pos - sMid 可以标注左子树的孩子个数即len.

AC Code:

<span style="font-size:14px;">/**
 * 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(vector<int> &inorder, vector<int> &postorder) {
        if(!inorder.size() || !postorder.size())
            return NULL;
        
        return buildTree_helper(inorder, postorder, 0, inorder.size()-1, 0, postorder.size()-1);
        
    }

private:
    TreeNode* buildTree_helper(vector<int>& inorder, vector<int>& postorder, int sMid, int eMid, int sPost, int ePost){
        if(sMid > eMid || sPost > ePost)
            return NULL;
        
        int poval = postorder[ePost];
        int pos;
        for(pos = sMid; pos <= eMid; pos++)
        {
            if(inorder[pos] == poval)
                break;
        } 
        
        TreeNode* root = new TreeNode(poval);
        int len = pos - sMid;
        root->left = buildTree_helper(inorder, postorder, sMid, pos-1, sPost,  sPost + len -1);
        root->right = buildTree_helper(inorder, postorder, pos + 1, eMid, sPost + len, ePost -1);
        
        return root;
    }
};</span>

Anwser 2: 非递归法

思路：根据后序遍历的特点，“左右根”；中序遍历，“左根右”；从后序遍历的末尾开始构造树，遇到的未和中序遍历匹配的点应该是根的右孩子，达到匹配时先遇到叶子结点的右孩子，回退（if(inorder.back() == vtn.back()->val)continue;）退回到根结点，然后构造左孩子。

Attention: 注意节点之间关系的构造以及何时push, pop.

AC Code:

<span style="font-size:14px;">/**
 * 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(vector<int> &inorder, vector<int> &postorder) {
        if(!inorder.size())
            return NULL;
            
        TreeNode* root, *p;
        vector<int> vint;
        vector<TreeNode*> vtn;
        
        root = new TreeNode(postorder.back());
        vtn.push_back(root);
        postorder.pop_back();
        
        while(true)
        {
            if(inorder.back() == vtn.back()->val)
            {
                p = vtn.back();
                vtn.pop_back();
                inorder.pop_back();
                if(inorder.size() == 0) break;
                //如果判断到右孩子，回退一个节点到根结点
                if(vtn.size())
                    if(inorder.back() == vtn.back()->val) continue;
                p->left = new TreeNode(postorder.back());
                postorder.pop_back();
                vtn.push_back(p->left);
            }
            else
            {
                p = new TreeNode(postorder.back());
                postorder.pop_back();
                vtn.back()->right = p;
                vtn.push_back(p);
            }
            
        }
        return root;
    }
};</span>