LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

Allenlzcoder

于 2018-06-24 16:12:04 发布

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分类专栏： LeetCode练习题文章标签： LeetCode 106. Construct Binary

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本文提供了一种解决LeetCode106题——从中间序列和后序序列构造二叉树的方法。通过递归地查找根节点并在中间序列中分割左右子树，实现了二叉树的构建。

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LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

Solution1:我的答案
仿照105题写的答案

/**
 * 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 {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return my_buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
    }

    TreeNode* my_buildTree(vector<int>& inorder, int iLeft, int iRight, vector<int>& postorder, int pLeft, int pRight) {
        if (iLeft > iRight || pLeft > pRight) return NULL;
        int i = 0;
        for (i = iLeft; i <= iRight; i++) {//2B的我写着2B的代码
            if (inorder[i] == postorder[pRight])
                break;
        }
        TreeNode* cur = new TreeNode(postorder[pRight]);
        cur->left = my_buildTree(inorder, iLeft, i - 1, postorder, pLeft, pLeft + i - iLeft - 1);
        cur->right = my_buildTree(inorder, i + 1, iRight, postorder, pLeft + i - iLeft, pRight - 1);
        return cur;

    }

};