【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal

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.

根据定义，后序遍历postorder的最后一个元素为根。

由于元素不重复，通过根可以讲中序遍历inorder划分为左子树和右子树。

递归下去即可求解。

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {

        return Helper(inorder, , inorder.size()-, postorder, , postorder.size()-);

    }

    TreeNode* Helper(vector<int>& inorder, int begin1, int end1, vector<int>& postorder, int begin2, int end2)

    {

        if(begin1 > end1)

            return NULL;

        else if(begin1 == end1)

            return new TreeNode(inorder[begin1]);

        TreeNode* root = new TreeNode(postorder[end2]);

        int i = begin1;

        for(; i <= end1; i ++)

        {

            if(inorder[i] == postorder[end2])

                break;

        }

        int leftlen = i-begin1;

        root->left = Helper(inorder, begin1, begin1+leftlen-, postorder, begin2, begin2+leftlen-);

        root->right = Helper(inorder, begin1+leftlen+, end1, postorder, begin2+leftlen, end2-);

        return root;

    }

};

巴特西

【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal

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