[leetcode] 105. Construct Binary Tree from Preorder and Inorder Traversal-优快云博客

本文介绍了一种使用前序遍历和中序遍历构建二叉树的方法。通过递归地在中序遍历中找到前序遍历的根节点来划分左右子树，并继续构建子树，直至所有节点都被处理。

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

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

解法一：

跟另一道几乎一样。对于preorder来说，第一个数对应着parent node。在inorder里找到这个数，然后recursive的分别对inorder的左右建tree。

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