Leetcode Construct Binary Tree from Preorder and Inorder Traversal

OpenSoucre發表於2014-07-02

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

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

此題目有兩種解決思路:

1)遞迴解決(比較好想)按照手動模擬的思路即可

2)非遞迴解決,用stack模擬遞迴

class Solution {
public:
    TreeNode *buildTree(vector<int>& preorder, int pre_left,int pre_right,
                        vector<int>& inorder,  int in_left, int in_right){
        if(pre_left > pre_right || in_left > in_right) return NULL;
        TreeNode *root = new TreeNode(preorder[pre_left]);
        int index = in_left;
        for( ; index <= in_right; ++ index ) if(inorder[index] == preorder[pre_left])  break;
        int left_cnt = index-in_left;
        root->left = buildTree(preorder,pre_left+1,pre_left+left_cnt,inorder,in_left,index-1);
        root->right = buildTree(preorder,pre_left+left_cnt+1,pre_right, inorder, index+1,in_right);
        return root;
    }
    
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size() == 0) return NULL;
        else return buildTree(preorder,0,preorder.size()-1, inorder,0,inorder.size()-1);
    }
};
遞迴解決

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