def rebuild(low, high):
if (high - low) > 0:
# find max val and idx
max_val = -1
for i in range(low, high):
if nums[i] > max_val: max_val,max_idx = (nums[i], i)
return TreeNode(val = max_val, left = rebuild(low, max_idx), right = rebuild(max_idx+1, high))
def rebuild(stop):
if inorder[-1] != stop:
root = TreeNode(val=preorder.pop())
root.left = rebuild(root.val)
inorder.pop()
root.right = rebuild(stop)
return root
def tree(height):
global i
if height:
root = TreeNode(val=i)
i += 1
root.left = tree(height-1)
root.right = tree(height-1)
return root
def rebuild(pl, ph, il, ih):
if pl >= ph: return None
cur = preorder[pl]
# search in inorder
idx = num2idx[cur]
n_left = idx-il # 左子树节点数
left = rebuild(pl+1, pl+1+n_left, il, idx)
right = rebuild(pl+1+n_left, ph, idx+1, ih)
return TreeNode(val=cur, left=left, right=right)
def rebuild(nums):
if len(nums) > 0:
# find max val and idx
max_val = -1
max_idx = -1
for i,x in enumerate(nums):
max_val,max_idx = (x, i) if x > max_val else (max_val, max_idx)
print(nums, max_val)
return TreeNode(val = max_val, left = rebuild(nums[:max_idx]), right = rebuild(nums[max_idx+1:]))
def rebuild(pl, ph, il, ih):
if il >= ih: return None
root = TreeNode(val = postorder[ph-1])
# search in indorder
idx = num2idx[root.val]
mid = ph-1-(ih-1-idx)
root.right = rebuild(mid, ph-1, idx+1, ih)
root.left = rebuild(pl, mid, il, idx)
return root
def rebuild(pl, ph, il, ih):
if pl >= ph: return None
cur = preorder[pl]
# search in inorder
for idx in range(il, ih):
if inorder[idx] == cur: break
n_left = idx-il # 左子树节点数
left = rebuild(pl+1, pl+1+n_left, il, idx)
right = rebuild(pl+1+n_left, ph, idx+1, ih)
return TreeNode(val=cur, left=left, right=right)
def rebuild(pl, ph, il, ih):
if il >= ih: return None
root = TreeNode(val = postorder[ph-1])
# search in indorder
for idx in range(il, ih):
if inorder[idx] == root.val: break
mid = ph-1-(ih-1-idx)
root.right = rebuild(mid, ph-1, idx+1, ih)
root.left = rebuild(pl, mid, il, idx)
return root
def rebuild(stop):
if inorder[-1] != stop:
# 新建当前节点
root = TreeNode(val=postorder.pop())
# 重构右子树，inoreder的stop为当前节点的值，找到则右子树重构完毕
root.right = rebuild(root.val)
# 注意右子树重构完毕，意味着inorder中已经不存在左子树的节点了
# 那么现在该干什么？当然是pop掉当前的节点了
inorder.pop()
# 重构左子树，左子树的stop值为本函数的stop相同，初始为None
root.left = rebuild(stop)
return root
def rebuild(low, high, flag):
if (high - low) > 0:
# find max val and idx
if flag:
max_idx= dp_idx[high-1]
max_val = nums[max_idx]
else:
max_val = -1
for i in range(low, high):
if nums[i] > max_val:
print(dp_idx[i])
max_val,max_idx = (nums[i], i)
dp_idx[i]= max_idx
return TreeNode(val = max_val, left = rebuild(low, max_idx, True), right = rebuild(max_idx+1, high, False))
def constructMaximumBinaryTree(self, nums):
stack = []
for num in nums:
# print(num, ': ', end=' ')
# for x in stack:
# print(x.val, end=' ')
# print()
cur = TreeNode(val=num)
# 找到比它小的做它左子节点
while stack and stack[-1].val < cur.val:
cur.left = stack.pop()
# 找到比它大的最小节点x，它作为x的右子节点
if stack:
stack[-1].right = cur
# cur称为右子树最末的节点，入栈
stack.append(cur)
return stack[0]
def tree(height):
if height:
root = TreeNode(val=0)
root.left = tree(height-1)
return root
def rebuild(low, high):
if (high - low) > 0:
# find max val and idx
return TreeNode(val = nums[dp_idx[low][high-1]], left = rebuild(low, dp_idx[low][high-1]), right = rebuild(dp_idx[low][high-1]+1, high))
def rebuild(nums):
if len(nums) > 0:
# find max val and idx
max_idx = nums.index(max(nums))
max_val = nums[max_idx]
return TreeNode(val = max_val, left = rebuild(nums[:max_idx]), right = rebuild(nums[max_idx+1:]))