從前序與中序遍歷序列構造二叉樹

根據一棵樹的前序遍歷與中序遍歷構造二叉樹。

注意:你可以假設樹中沒有重複的元素。

例如，給出

前序遍歷 preorder = [3,9,20,15,7]

中序遍歷 inorder = [9,3,15,20,7]

3

/ \9  20
/  \
15   7

# definition for a binary tree node.

# class treenode:
#     def __init__(self, x):
#         self.val = x
#         self.left = none
#         self.right = none
class
solution
:def
buildtree
(self, preorder, inorder)
:"""
:type preorder: list[int]
:type inorder: list[int]
:rtype: treenode
"""defhelper
(in_left =
0, in_right =
len(inorder)):
nonlocal pre_idx
# if there is no elements to construct subtrees
if in_left == in_right:
return
none
# pick up pre_idx element as a root
root_val = preorder[pre_idx]
root = treenode(root_val)
# root splits inorder list
# into left and right subtrees
index = idx_map[root_val]
# recursion 
pre_idx +=
1# build left subtree
root.left = helper(in_left, index)
# build right subtree
root.right = helper(index +
1, in_right)
return root
# start from first preorder element
pre_idx =
0# build a hashmap value -> its index
idx_map =
return helper(
)

/**

* definition for a binary tree node.
* struct treenode 
* };
*/class
solution
treenode*
buildtree
(vector<
int>
& preorder, vector<
int>
& inorder)
};

從前序與中序遍歷序列構造二叉樹

題目描述 if inbegin inend 區間只有乙個結點，就是根結點 區間正常 int rootindex inbegin while rootindex inend 用前序的根劃分中序為兩個子區間 else 遞迴建立左子樹 root left buildtree preorder,pindex...

從前序與中序遍歷序列構造二叉樹

根據一棵樹的前序遍歷與中序遍歷構造二叉樹。definition for a binary tree node.public class treenode class solution int rootidx left while rootidx right rootidx 前序 3 9 20 15 ...

從前序與中序遍歷序列構造二叉樹

根據一棵樹的前序遍歷與中序遍歷構造二叉樹。解題思路 left,right 這個區間就表示當前preorder index 這個結點對應的子樹的中序遍歷結果 private treenode buildtreehelper int preorder,int inorder,int left,int r...