(二叉树 递归) leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7
-----------------------------------------------------------------------------------
就是从先序遍历和中序遍历构建踹个二叉树，可以用递归方式，注意递归的终止条件。
和 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal几乎一样。

参考博客：http://www.cnblogs.com/grandyang/p/4296500.html

C++代码：

/**
 * 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) {
        return build(preorder,0,preorder.size() - 1,inorder,0,inorder.size() - 1);
    }
    TreeNode* build(vector<int> &preorder,int pleft,int pright,vector<int> &inorder,int ileft,int iright){
        if(pleft > pright || ileft > iright) return NULL;  //递归的终止条件。
        int i = 0;
        TreeNode *cur = new TreeNode(preorder[pleft]);
        for(i = ileft; i < inorder.size(); i++){
            if(inorder[i] == cur->val)
                break;
        }
        cur->left = build(preorder,pleft + 1,pleft + i - ileft,inorder,ileft,i-1);
        cur->right = build(preorder,pleft + i - ileft + 1,pright,inorder,i+1,iright);
        return cur; 
    }
};

 也有一个方法：

/**
 * 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) {
        return build(preorder,inorder);
    }
    TreeNode* build(vector<int> &preorder,vector<int> &inorder){
        if(preorder.size() == 0 || inorder.size() == 0) return NULL;  //递归条件。当然，也还可以加上if(preorder.size() == 1 || inorder.size() == 1) return cur;这个递归条件。
        int rootval = preorder[0];
        int i = 0;
        for(i = 0; i < inorder.size(); i++){
            if(inorder[i] == rootval)
                break;
        }
        vector<int> inleft,inright,preleft,preright;
        for(int k = 1; k < i + 1; k++)
            preleft.push_back(preorder[k]);
        for(int k = i + 1; k < preorder.size(); k++){
            preright.push_back(preorder[k]);
            inright.push_back(inorder[k]);
        }
        for(int k = 0; k < i; k++)
            inleft.push_back(inorder[k]);
        TreeNode *cur = new TreeNode(rootval);
        cur->left = build(preleft,inleft);
        cur->right = build(preright,inright);
        return cur;
    }
};

这两个方法从思想上是一样的，只不过代码的实现有所不同。