[LeetCode] 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.

解题思路：

可以用递归的办法做。由后序遍历确定根节点，有中序遍历确定左右子树。每个子树有分别是一个后序遍历和一个中序遍历。代码如下。注意递归调用时位置的计算，后序遍历和中序遍历的长度一致。

/** * 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 getTreeNode(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size()-1);    }        TreeNode *getTreeNode(vector<int> &inorder, int startInOrder, int endInOrder,                           vector<int> & postorder, int startPostOrder, int endPostOrder){        if(startInOrder == endInOrder){            return new TreeNode(postorder[startPostOrder]);        }else if(startInOrder > endInOrder){            return NULL;        }else{            int postNumber = postorder[endPostOrder];            //找到postNumber在中序遍历的位置            int inPosition=0;            for(int i=startInOrder; i<=endInOrder; i++){                if(inorder[i]==postNumber){                    inPosition=i;                    break;                }            }            TreeNode* node = new TreeNode(postNumber);            node->right = getTreeNode(inorder, inPosition+1,endInOrder, postorder, endPostOrder+inPosition-endInOrder, endPostOrder - 1);            node->left = getTreeNode(inorder, startInOrder, inPosition-1, postorder, startPostOrder, inPosition - 1 - startInOrder + startPostOrder);            return node;        }    }};