力扣 面试题07. 重建二叉树 递归分治/非递归版

https://leetcode-cn.com/problems/zhong-jian-er-cha-shu-lcof/

思路：先序遍历：根左右。中序遍历：左根右。那么可以由先序遍历得到根节点的值，依据此值再找到根节点在中序遍历的位置，那么就可以分出左子树和右子树，分治下去即可。

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

    TreeNode* dfs(int l1,int r1,int l2,int r2,vector<int>& preorder,vector<int>& inorder)
    {
        if(l1==r1)//只有一个节点时
        {
            TreeNode *node=new TreeNode(preorder[l1]);
            return node;
        }
        int rootval=preorder[l1];//根节点的值
        int i=l2;
        while(i<=r2&&inorder[i]!=rootval)//在中序遍历中找到根节点
            ++i;
        int lnum=i-l2;//左子树元素个数
        int rnum=r2-i;//右子树元素个数
        TreeNode *root=new TreeNode(rootval);
        TreeNode *lc=NULL,*rc=NULL;
        if(lnum)
            lc=dfs(l1+1,l1+lnum,l2,i-1,preorder,inorder);
        if(rnum)
            rc=dfs(r1-rnum+1,r1,i+1,r2,preorder,inorder);
        root->left=lc;
        root->right=rc;
        return root;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int siz=preorder.size();
        if(!siz)
            return NULL;
        return dfs(0,siz-1,0,siz-1,preorder,inorder);
    }
};

非递归版？感觉这个有点难理解，最好结合例子体会一下。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int siz=preorder.size();
        if(!siz)
            return NULL;
        stack<TreeNode*> s;
        int j=0;
        TreeNode *node,*head=new TreeNode(preorder[0]);
        s.push(head);
        for(int i=1;i<siz;i++)
        {
            if(s.top()->val!=inorder[j])
            {
                node=new TreeNode(preorder[i]);
                s.top()->left=node;
                s.push(node);
            }
            else
            {
                while(!s.empty()&&s.top()->val==inorder[j])
                    node=s.top(),s.pop(),j++;
                TreeNode *tmp=new TreeNode(preorder[i]);
                node->right=tmp;
                s.push(tmp);
            }
        }
        return head;
    }
};