leetcode 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.

遇到二叉树问题最先想到的是递归解决。

使用 TreeNode * 引用，来保存暂时结果

/**
 * 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) {
        TreeNode *root=NULL;
        if(inorder.empty()) return root;
        sub(root,inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
        return root;
        
    }
    void sub(TreeNode *&root,vector<int> &inorder,int istart,int iend,vector<int> &postorder,int pstart,int pend){
        if(istart>iend||pstart>pend){
            root=NULL;
            return;
        }
        root=new TreeNode(postorder[pend]);
        int split;
        for(int i=istart;i<=iend;i++){
            if(inorder[i]==postorder[pend]){
                split=i;
            }
        }
        int leftnum=split-istart;  //表示左子树的节点个数
        TreeNode *l,*r;
        sub(l,inorder,istart,istart+leftnum-1,postorder,pstart,pstart+leftnum-1);
        sub(r,inorder,istart+leftnum+1,iend,postorder,pstart+leftnum,pend-1);
        root->left=l;
        root->right=r;
    }
};

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