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Construct Binary Tree from Preorder and Inorder Traversal

 
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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.

 

 

// construct tree from preorder and inorder
 
class Solution {
    
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size() == 0) return NULL;
        
        return buildTree(preorder, 0, inorder, 0, preorder.size());
    }
    
    TreeNode *buildTree(vector<int> &pre, int preStart, vector<int> &in, int inStart, int len) {
        if(preStart <  0 || preStart + len > pre.size() || inStart < 0 || inStart + len > in.size() || len < 1) {
            return NULL;
        }
        
        int rootVal = pre[preStart];
        
        int i = 0;
        while(in[i] != rootVal) ++i;
        
        TreeNode *root = new TreeNode(rootVal);
        
        int leftLen = i - inStart;
        int rightLen = len - leftLen - 1;
        
        root->left = buildTree(pre, preStart+1, in, inStart, leftLen);
        root->right = buildTree(pre, preStart+leftLen+1, in, i+1, rightLen);
        
        return root;
    }
};

 

 

 

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