leetcode:Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2022-01-01 11:35:53 发布

原创最新推荐文章于 2022-01-01 11:35:53 发布 · 328 阅读

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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。该方法首先确定根节点，然后递归地构建左子树和右子树。文章提供了一个高效的C++实现示例。

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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 *buildTree(vector<int> &inorder, vector<int> &postorder) {
        
        if (inorder.size()==0 && postorder.size()==0)
            return NULL;
        
        int pivot = postorder[postorder.size()-1];
        TreeNode *root = new TreeNode(pivot);
        
        vector<int> left_inorder;
        vector<int> left_postorder;
        
        vector<int> right_inorder;
        vector<int> right_postorder;
        
        int i = 0;
        while (inorder[i]!=pivot && i<inorder.size())
        {
            left_inorder.push_back(inorder[i]);
            left_postorder.push_back(postorder[i]);
            i++;
        }
        
        for (int j=i+1; j<inorder.size(); j++)
            right_inorder.push_back(inorder[j]);
            
        for (int j=i; j<postorder.size()-1; j++)
            right_postorder.push_back(postorder[j]);
            
        root->left = buildTree(left_inorder, left_postorder);
        root->right = buildTree(right_inorder, right_postorder);
        
        return root;
    }
*/

    TreeNode *helper(vector<int> &inorder, int i_startIdx, int i_endIdx, vector<int> &postorder, int p_startIdx, int p_endIdx) {
        
        if (i_endIdx<i_startIdx || p_endIdx<p_startIdx)
            return NULL;
    
        int pivot = postorder[p_endIdx];
        TreeNode *root = new TreeNode(pivot);
    
        int pivotIdx = i_startIdx;
        while (inorder[pivotIdx]!=pivot && pivotIdx<=i_endIdx)
        {
            pivotIdx++;
        }
    
        root->left  = helper(inorder, i_startIdx, pivotIdx-1, postorder, p_startIdx, p_startIdx+(pivotIdx-1-i_startIdx));
        root->right = helper(inorder, pivotIdx+1, i_endIdx, postorder, p_startIdx+(pivotIdx-i_startIdx), p_endIdx-1);
    
        return root;
    }
    
    
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        
        if (inorder.size()==0 && postorder.size()==0)
            return NULL;
            
        return helper(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);
    }
};