LeetCode: 106. Construct Binary Tree from Inorder and Postorder Traver

题目描述

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

    3
   / \   9  20
    /  \    15   7

• 题目大意: 根据二叉树的中序遍历和后序遍历，构造该二叉树。

解题思路

与 LeetCode: 105. Construct Binary Tree from Preorder and Inorder Traversal 题解 的方法类似。 根据后序遍历序列，找到根节点，然后根据根节点在中序遍历序列中的位置，拆分出左子树和右子树。分别对左子树，右子树做同样的操作，直到叶节点。

AC 代码

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution 
{
private:
    TreeNode* buildTree(vector<int>& inorder, int inBeg, 
                       vector<int>& postorder, int postBeg, int size)
    {
        if(size <= 0) return nullptr;

        TreeNode* curNode = new TreeNode(postorder[postBeg+size-1]);

        int inorderLeftTreeEnd = inBeg;
        while(inorder[inorderLeftTreeEnd] != curNode->val) ++inorderLeftTreeEnd;

        int leftTreeSize = inorderLeftTreeEnd - inBeg;
        int rightTreeSize = (inBeg + size) - inorderLeftTreeEnd - 1;

        curNode->left = buildTree(inorder, inBeg, postorder, postBeg, leftTreeSize);
        curNode->right = buildTree(inorder, inorderLeftTreeEnd + 1, 
                                  postorder, postBeg+leftTreeSize, rightTreeSize);

        return curNode;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) 
    {
        return buildTree(inorder, 0, postorder, 0, inorder.size());
    }
};