Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2021-02-03 15:56:42 发布

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本文介绍了一种通过给定的中序遍历和后序遍历序列来构建二叉树的方法。该方法首先创建一个映射表以快速查找节点的位置，然后递归地构建树的左右子树。

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

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

/**
 * 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* visit(vector<int> &inorder, int lbegin, int lend, vector<int> &postorder, int rbegin, int rend, map<int, int> &table)
    {
        if (lbegin > lend)
        {
            return NULL;
        }
        TreeNode *root = new TreeNode(postorder[rend]);
        int pos = table[postorder[rend]];
        int offset = pos - lbegin - 1;
        root->left = visit(inorder, lbegin, pos-1, postorder, rbegin, rbegin+offset, table);
        root->right = visit(inorder, pos+1, lend, postorder, rbegin+offset+1, rend-1, table);
        return root;
    }
        
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        int size = inorder.size();
        if (size < 1)
        {
            return NULL;
        }
    
        map<int, int> table;
        for (int i = 0; i < size; i++)
        {
            table[inorder[i]] = i;
        }

        return visit(inorder, 0, size-1, postorder, 0, size-1, table);
    }
};