具有所有最深节点的最小子树

一开始题没看懂，他这里就是找到最深的叶子结点，看他们附近有没有公共祖先，如果有的话就意味着，这两个最深的叶子结点的深度是一样的，直接出他们附近的root，如果没有的话那其实就是看是不是他自己了。

class Solution:def lcaDeepestLeaves(self, root: Optional[TreeNode]) -> Optional[TreeNode]:def depth(root):if not root:return 0left = depth(root.left)right = depth(root.right)return max(left,right) + 1if not root:return None left = depth(root.left)right = depth(root.right)if left == right:return root elif left > right:return self.lcaDeepestLeaves(root.left)else:return self.lcaDeepestLeaves(root.right)

二叉树中的最长交错路径

这道题真的很难，就是不知道怎么控制方向，所以我们用了两个数字做result，一个代表接着往左走还有几个，另一个代表接着往右走还有几个，这里，如果第一步选择了root.left，那接着应该往右就是r1+1，如果第一步选择了root.right，那接着应该往左就是l2+1，往上想一步，root.left的上一层应该是属于再往上那颗的右节点，所以我们dfs(root.left)的想要的答案应该是那个右节点继续往左走的答案也就是l2+1，所以我们的return第二位的就是l2+1，反过来是一样的。

class Solution:def longestZigZag(self, root: Optional[TreeNode]) -> int:res = 0def dfs(root):nonlocal resif not root:return -1, -1l1, r1 = dfs(root.left)l2, r2 = dfs(root.right)res = max(res,1+r1,l2+1)return 1+r1, l2+1dfs(root)return res

最小体力消耗路径

例子都对了但提交的时候超时了。

class Solution:def minimumEffortPath(self, heights: List[List[int]]) -> int:row = len(heights)col = len(heights[0])res = []result = []used = [[False]*col for _ in range(row)]used[0][0] = Truedef dfs(i,j):nonlocal resif i == row-1 and j == col-1:result.append(max(res))return for k1,k2 in [[i+1,j],[i-1,j],[i,j+1],[i,j-1]]:if 0 <= k1 < row and 0 <= k2 < col and not used[k1][k2]:if result and min(result) < abs(heights[i][j]-heights[k1][k2]): continue#if res and max(res) < abs(heights[i][j]-heights[k1][k2]): continueres.append(abs(heights[i][j]-heights[k1][k2]))used[k1][k2] = Truedfs(k1,k2)used[k1][k2] = Falseres.pop()if row == col == 1:return 0dfs(0,0)return min(result)

一个并查集的例子做到了再回来看。

class Solution:def minimumEffortPath(self, heights: List[List[int]]) -> int:m = len(heights)n = len(heights[0])fa = [i for i in range(m*n)]def find(x):if x == fa[x]:return xelse:fa[x] = find(fa[x])return fa[x]edges = []for i in range(m):for j in range(n):to = i * n + jif i > 0:edges.append((to - n,to,abs(heights[i][j] - heights[i - 1][j])))if j > 0:edges.append((to - 1,to,abs(heights[i][j] - heights[i][j - 1])))edges.sort(key=lambda x:x[2])ans = 0for x,y,v in edges:fx,fy = find(x),find(y)fa[fy] = fxfs = find(0)fe = find(m*n - 1)if fs == fe:ans = vbreakreturn ans