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  • 重构二叉树

    重构二叉树的思路主要是首先在前序(后序)序列中找到根结点,然后在中序序列中找到根结点所在的位置,该结点将整个序列分成两个部分,前一部分为根结点的左子树元素,后一部分为根结点的右子树元素。再递归的生成左子树和右子树即可。

    1.通过前序,中序序列重构二叉树

    TreeNode * buildTree(vector<int>& preorder, int preBegin, int preEnd, vector<int>& inorder, int inBegin, int inEnd)
     {
        TreeNode*root =NULL;
    
        if (preEnd >= preBegin)
        {
            root = new TreeNode(preorder[preBegin]);
            int rootIndex = 0;
            for (int i = inBegin; i <= inEnd; ++i)
            {
                if (inorder[i] == preorder[preBegin])
                {
                    rootIndex = i;
                    break;
                }
            }
    
            if (rootIndex!=inBegin)
            {
                root->left = buildTree(preorder, preBegin+1, preBegin+(rootIndex-inBegin), inorder, inBegin, rootIndex-1);
            }
            if (rootIndex!=inEnd)
            {
                root->right = buildTree(preorder, preBegin+1+(rootIndex-inBegin), preEnd, inorder, rootIndex+1, inEnd);
            }
        }
        return root;  
     }
        TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
            return buildTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
        }

    2.通过中序,后序序列重构二叉树

    TreeNode * buildTree(vector<int>& inorder, int inBegin, int inEnd, vector<int>& postorder, int postBegin, int postEnd)
     {
        TreeNode*root =NULL;
    
        if (postEnd >= postBegin)
        {
            root = new TreeNode(postorder[postEnd]);
    
            int rootIndex = 0;
            for (int i = inBegin; i <= inEnd; ++i)
            {
                if (inorder[i] == postorder[postEnd])
                {
                    rootIndex = i;
                    break;
                }
            }
    
            if (rootIndex!=inBegin)
            {
                root->left = buildTree( inorder, inBegin, rootIndex-1, postorder, postBegin, postBegin+(rootIndex-inBegin)-1);
            }
            if (rootIndex!=inEnd)
            {
                root->right = buildTree(inorder, rootIndex+1, inEnd, postorder, postBegin+(rootIndex-inBegin), postEnd-1);
            }
        }
        return root;  
     }
        TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
          return buildTree(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);  
        }
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  • 原文地址:https://www.cnblogs.com/happygirl-zjj/p/4574616.html
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