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  • leetcode -day23 Construct Binary Tree from Inorder and Postorder Traversal & Construct Binary Tree f

    1、
    

    Construct Binary Tree from Inorder and Postorder Traversal

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

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

    代码:

    class Solution {
    public:
        TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
            TreeNode* root = NULL;
            int len1 = inorder.size();
            int len2 = postorder.size();
            if(len1<1 || len2<1 || len1!=len2){
                return root;
            }
            return buildTreeCore(inorder,0,len1-1,postorder,0,len2-1);
        }
        TreeNode* buildTreeCore(vector<int>&inorder, int startIndex1, int endIndex1, vector<int>&postorder, int startIndex2, int endIndex2){
            if(endIndex1 < startIndex1  ){
                return NULL;
            }
            TreeNode* root = new TreeNode(postorder[endIndex2]);
            int index = 0;
            for(int i=startIndex1; i<=endIndex1; ++i){
                if(inorder[i] == postorder[endIndex2]){
                    index = i;
                    break;
                }
            }
            int leftLen = index-startIndex1;
            TreeNode* leftNode = NULL;
            TreeNode* rightNode = NULL;
           
            leftNode = buildTreeCore(inorder,startIndex1,index-1,postorder,startIndex2,startIndex2+leftLen-1);
            rightNode = buildTreeCore(inorder,index+1,endIndex1,postorder,startIndex2+leftLen,endIndex2-1);
          
            root->left = leftNode;
            root->right = rightNode;
            return root;
        }
    };

    2、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.

    class Solution {
    public:
        TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
            TreeNode* root = NULL;
            int len1 = inorder.size();
            int len2 = preorder.size();
            if(len1<1 || len2<1 || len1!=len2){
                return root;
            }
            return buildTreeCore(inorder,0,len1-1,preorder,0,len2-1);
        }
          TreeNode* buildTreeCore(vector<int>&inorder, int startIndex1, int endIndex1, vector<int>&preorder, int startIndex2, int endIndex2){
            if(endIndex1 < startIndex1  ){
                return NULL;
            }
            TreeNode* root = new TreeNode(preorder[startIndex2]);
            int index = 0;
            for(int i=startIndex1; i<=endIndex1; ++i){
                if(inorder[i] == preorder[startIndex2]){
                    index = i;
                    break;
                }
            }
            int leftLen = index-startIndex1;
            TreeNode* leftNode = NULL;
            TreeNode* rightNode = NULL;
           
            leftNode = buildTreeCore(inorder,startIndex1,index-1,preorder,startIndex2+1,startIndex2+leftLen);
            rightNode = buildTreeCore(inorder,index+1,endIndex1,preorder,startIndex2+leftLen+1,endIndex2);
          
            root->left = leftNode;
            root->right = rightNode;
            return root;
        }
    };


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  • 原文地址:https://www.cnblogs.com/jzssuanfa/p/7066980.html
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