【二叉树经典问题】106. Construct Binary Tree from Inorder and Postorder Traversal

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 a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */

解答：

题目要求通过中序（左、根、右）和后序（左、右、根）重建一棵二叉树，例如

/*例：              1    /\   23   /\   4 5后序：4 5 2 3 1中序：4 2 5 1 3*/

构建的方法其实很简单：后序的最后一个元素是根，在中序中找到它，那么按照中序的特点，这个元素左边的就是左子树，右边的就是右子树，这样递归做下去就可以了。

代码：

class Solution {public:    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {        return dfs(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);    }    TreeNode* dfs(vector<int>& inorder,vector<int>& postorder,int is,int ie,int ps,int pe){        if(is>ie)            return nullptr;        TreeNode* root=new TreeNode(postorder[pe]);        int pos;        for(int i=is;i<=ie;i++){            if(inorder[i]==root->val){                pos=i;                break;            }        }        root->left=dfs(inorder,postorder,is,pos-1,ps,ps+pos-is-1);//because pe-ps=pos-1-is (the number of the rest elements should be equal)        root->right=dfs(inorder,postorder,pos+1,ie,pe-ie+pos,pe-1);//pe-1-ps=ie-(pos+1) => ps=pe-ie+pos        return root;    }};

说起来。。有想法和能把想法转换成代码这之间是有一定差距的。。。还有，看了别人的代码就“懂”了，自己写却不会，这是最悲伤的quq

这个问题还有非递归的解法，代码来自Discuss：

using namespace std;#define forLess(i, s, e)for(int i = (s); i < int(e); ++i)#define forto(i, s, e)for(int i = (s); i <= int(e); ++i)class Solution {public:     TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)     {    TreeNode* cur = NULL;    int i = 0, j = 0;    stack<TreeNode *> s;    while(i < inorder.size() && j < postorder.size()){if(inorder[i] == postorder[j]){    TreeNode *t = new TreeNode(inorder[i]);    t->left = cur;    cur = t;    i++;    j++;    }    else{    if(!s.empty() && postorder[j] == s.top()->val){    TreeNode *t = s.top();    s.pop();    t->right = cur;    cur = t;    j++;    }    else{    TreeNode *t = new TreeNode(inorder[i]);    t->left = cur;    cur = NULL;    s.push(t);    i++;    }    }    }    while(!s.empty()){    TreeNode *t = s.top();    s.pop();    t->right = cur;    cur = t;    }    return cur;    }};