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
這道題要求用先序和中序遍歷來建立二叉樹,跟之前那道 Construct Binary Tree from Inorder and Postorder Traversal 原理基本相同,針對這道題,由於先序的順序的第一個肯定是根,所以原二叉樹的根節點可以知道,題目中給了一個很關鍵的條件就是樹中沒有相同元素,有了這個條件就可以在中序遍歷中也定位出根節點的位置,並以根節點的位置將中序遍歷拆分為左右兩個部分,分別對其遞歸調用原函數,參見代碼如下:
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; } };
下面來看一個例子, 某一二叉樹的中序和后序遍歷分別為:
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
做完這道題后,大多人可能會有個疑問,怎么沒有由先序和后序遍歷建立二叉樹呢,這是因為先序和后序遍歷不能唯一的確定一個二叉樹,比如下面五棵樹:
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
從上面我們可以看出,對於先序遍歷都為 1 2 3 的五棵二叉樹,它們的中序遍歷都不相同,而它們的后序遍歷卻有相同的,所以只有和中序遍歷一起才能唯一的確定一棵二叉樹。但可能會有小伙伴指出,那第 889 題 Construct Binary Tree from Preorder and Postorder Traversal 不就是從先序和后序重建二叉樹么?難道博主被啪啪打臉了么?難道博主的一世英名就此毀於一旦了么?不,博主向命運的不公說不,請仔細看那道題的要求 "Return any binary tree that matches the given preorder and postorder traversals.",是讓返回任意一棵二叉樹即可,所以這跟博主的結論並不矛盾。長舒一口氣,博主的晚節保住了~
Github 同步地址:
https://github.com/grandyang/leetcode/issues/105
類似題目:
Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Preorder and Postorder Traversal
參考資料:
https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/