LeetCode 889. Construct Binary Tree from Preorder and Postorder Traversal

最新推荐文章于 2020-03-01 11:14:51 发布

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本文介绍了一种方法，通过给定的预序和后序遍历序列来构建二叉树。该方法确保了即使存在多个可能的解时也能返回其中之一。文章详细解释了构建过程，并提供了具体的实现代码。

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Return any binary tree that matches the given preorder and postorder traversals.

Values in the traversals pre and post are distinct positive integers.

Example 1:

Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]

Note:

1 <= pre.length == post.length <= 30
pre[] and post[] are both permutations of 1, 2, ..., pre.length.
It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.

/**
 * 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* tree(vector<int>& pre, vector<int>& post,int pa,int pb,int ia,int ib){
        if(pb<pa)
            return NULL;
        TreeNode *t=new TreeNode(pre[pa]);
        if(pb==pa)
            return t;
        int k;
        for(k=ia;k<ib;++k)
            if(post[k]==pre[pa+1])
                break;
        t->left=tree(pre,post,pa+1,pa+k-ia+1,ia,k);
        t->right=tree(pre,post,pa+k-ia+2,pb,k+1,ib-1);
        return t;
    }
    TreeNode* constructFromPrePost(vector<int>& pre, vector<int>& post) {
         int n=pre.size();
        if(n==0)
            return NULL;
        TreeNode *t=tree(pre,post,0,n-1,0,n-1);
        return t;
    }
};