版权声明：QQ:872289455. WeChat:luw9527. https://blog.csdn.net/MC_007/article/details/80849136
105. Construct Binary Tree from Preorder and Inorder Traversal
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
画了一个复杂点的二叉树来剖析本题的递归，注意尾递归不在搜索范围内，只是用来确定边界，边界一定要考虑清除！
class
Solution {
public:
TreeNode
*
buildTree(vector
<
int
>& preorder, vector
<
int
>& inorder) {
return
buildTree(
begin(preorder),
end(preorder),
begin(inorder),
end(inorder));
}
template
<
typename T
>
TreeNode
*
buildTree(T pre_first, T pre_last,T in_first, T in_last) {
if (pre_first
== pre_last
|| in_first
== in_last)
return
nullptr;
auto root
=
new
TreeNode(
*pre_first);
auto RootPos
=
find(in_first, in_last, root->
val);
auto leftSize
=
distance(in_first, RootPos);
auto boundary
=
next(pre_first, leftSize
+
1);
root->
left
=
buildTree(
next(pre_first), boundary, in_first, RootPos);
root->
right
=
buildTree(boundary, pre_last,
next(RootPos), in_last);
return root;
}
};
106. Construct Binary Tree from Inorder and Postorder Traversal
given
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
同样以上面的图为例来说明：
class
Solution {
public:
TreeNode
*
buildTree(vector
<
int
>& inorder, vector
<
int
>& postorder) {
return
buildTree(
begin(inorder),
end(inorder),
begin(postorder),
end(postorder));
}
template
<
typename T
>
TreeNode
*
buildTree(T in_first, T in_last, T post_first, T post_last) {
if (in_first
== in_last
|| post_first
== post_last)
return
nullptr;
const
int val
=
*
prev(post_last);
TreeNode
*root
=
new
TreeNode(val);
auto rootPos
=
find(in_first, in_last, val);
auto leftsize
=
distance(in_first, rootPos);
auto boundary
=
next(post_first, leftsize);
root->
left
=
buildTree(in_first, rootPos, post_first, boundary);
root->
right
=
buildTree(
next(rootPos), in_last, boundary,
prev(post_last));
return root;
}
};