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leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

时间:2017-08-19 11:05:37      阅读:130      评论:0      收藏:0      [点我收藏+]

标签:遍历   solution   index   中序遍历   dex   序列   先序遍历   new   相对   

递归解法:

从先序遍历中得到树的根节点,从中序遍历中得到左右数的组成节点。

技术分享

 

class Solution {
public:
      TreeNode *creTree(vector<int> preorder, int i,int j, vector<int> inorder,int m, int n)
        {
         
         // cout<<i<<"  "<<j<<endl;
          if(i>j&& m>n)
          {
               return NULL;
          }
             
        
       //得到根节点
       TreeNode * t=new TreeNode(preorder[i]);
    
          int k;
        for( k=m;k<n;k++)
            if(preorder[i]==inorder[k])
                break;

//分开左右子树
//这里是 i+k-m, 不是i+k, 因为k是在整个序列中的位置,而不是相对于m的位置 t->left=creTree(preorder,i+1, i+k-m,inorder, m,k-1); t->right=creTree(preorder, i+k-m+1, j,inorder, k+1, n ); return t; } TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { if(preorder.size()==0) return NULL; //cout<<* preorder.begin()<<endl; //cout<< * preorder.end()<<endl; // if(preorder.begin()==preorder.end()-1) // cout<<"ok"<<endl; return creTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1); } };

 更好理解的解法:

public TreeNode build(int[] preorder, int[] inorder, int preIndex,
            int startInIndex, int endInIndex) {
        if (endInIndex < startInIndex) {
            return null;
        }
        TreeNode node = new TreeNode(preorder[preIndex]);

        // the index of current node in inorder
        int index = getIndexInInorder(inorder, preorder[preIndex]);
        int lenL = index - startInIndex;
        int lenR = endInIndex - startInIndex - lenL;

        if (lenL > 0) {
            node.left = build(preorder, inorder, preIndex + 1, startInIndex,
                    index - 1);
        }
        if (lenR > 0) {
            node.right = build(preorder, inorder, preIndex + lenL + 1,
                    index + 1, endInIndex);
        }

        return node;
    }

  

public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        if (inorder == null || inorder.length == 0) {
            return null;
        }

        if (preorder.length != inorder.length) {
            return null;
        }

        return build(preorder, inorder, 0, 0, inorder.length - 1);
    }

  

 

leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

标签:遍历   solution   index   中序遍历   dex   序列   先序遍历   new   相对   

原文地址:http://www.cnblogs.com/fanhaha/p/7393285.html

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