【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
		    /	\
		   2	3
		   /\
		   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;
    }
};