105. Construct Binary Tree from Preorder and Inorder Traversal

 

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

 

这道题要求用先序和中序遍历来建立二叉树,跟之前那道 Construct Binary Tree from Inorder and Postorder Traversal 原理基本相同,针对这道题,由于先序的顺序的第一个肯定是根,所以原二叉树的根节点可以知道,题目中给了一个很关键的条件就是树中没有相同元素,有了这个条件就可以在中序遍历中也定位出根节点的位置,并以根节点的位置将中序遍历拆分为左右两个部分,分别对其递归调用原函数,参见代码如下:

 

class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        return buildTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
    TreeNode *buildTree(vector<int> &preorder, int pLeft, int pRight, vector<int> &inorder, int iLeft, int iRight) {
        if (pLeft > pRight || iLeft > iRight) return NULL;
        int i = 0;
        for (i = iLeft; i <= iRight; ++i) {
            if (preorder[pLeft] == inorder[i]) break;
        }
        TreeNode *cur = new TreeNode(preorder[pLeft]);
        cur->left = buildTree(preorder, pLeft + 1, pLeft + i - iLeft, inorder, iLeft, i - 1);
        cur->right = buildTree(preorder, pLeft + i - iLeft + 1, pRight, inorder, i + 1, iRight);
        return cur;
    }
};

 

下面来看一个例子, 某一二叉树的中序和后序遍历分别为:

Preorder:    5  4  11  8  13  9

Inorder:    11  4  5  13  8  9

 

5  4  11  8  13  9      =>          5

11  4  5  13  8  9                /  \

 

4  11     8   13  9      =>         5

11  4     13  8  9                  /  \

                             4   8

 

11       13    9        =>         5

11       13    9                    /  \

                             4   8

                            /    /     \

                           11    13    9

 

做完这道题后,大多人可能会有个疑问,怎么没有由先序和后序遍历建立二叉树呢,这是因为先序和后序遍历不能唯一的确定一个二叉树,比如下面五棵树:

    1      preorder:    1  2  3
   / \       inorder:       2  1  3
 2    3       postorder:   2  3  1

 

       1       preorder:     1  2  3
      /       inorder:       3  2  1
    2          postorder:   3  2  1
   /
 3

 

       1        preorder:    1  2  3
      /        inorder:      2  3  1
    2       postorder:  3  2  1
      \
       3

 

       1         preorder:    1  2  3
         \        inorder:      1  3  2
          2      postorder:  3  2  1
         /
       3

 

       1         preorder:    1  2  3
         \      inorder:      1  2  3
          2      postorder:  3  2  1
            \
    3

 

从上面我们可以看出,对于先序遍历都为 1 2 3 的五棵二叉树,它们的中序遍历都不相同,而它们的后序遍历却有相同的,所以只有和中序遍历一起才能唯一的确定一棵二叉树。但可能会有小伙伴指出,那第 889 题 Construct Binary Tree from Preorder and Postorder Traversal 不就是从先序和后序重建二叉树么?难道博主被啪啪打脸了么?难道博主的一世英名就此毁于一旦了么?不,博主向命运的不公说不,请仔细看那道题的要求 “Return any binary tree that matches the given preorder and postorder traversals.”,是让返回任意一棵二叉树即可,所以这跟博主的结论并不矛盾。长舒一口气,博主的晚节保住了~

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/105

 

类似题目:

Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Preorder and Postorder Traversal

 

参考资料:

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/discuss/34538/My-Accepted-Java-Solution

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/discuss/34562/Sharing-my-straightforward-recursive-solution

 

LeetCode All in One 题目讲解汇总(持续更新中…) 


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