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# liyunrui/1740. Find Distance in a Binary Tree.py

Created Feb 17, 2021
leetcode-tree
 # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: """ This code reuses the recursive function from Lowest Common Ancestor of Binary Tree (#236) Step1: We want to find the LCA since it is guaranteed to be on the path between p and q. Since we're unable to move "up" in a binary tree, we can simple move down in both directions from the LCA to form the path. Step2: Return the distance from the LCA -> p, plus the distance from the LCA -> q. Time complexity is O(n) since we have to look at all the nodes at least once in the worst case. Space complexity is O(logn) for the call stack. Also, this question shouldn't be a medium considering that #236 is a medium itself lol. Ref: https://leetcode.com/problems/find-distance-in-a-binary-tree/discuss/1039730/Clean-Python-using-Lowest-Common-Ancestor """ def findDistance(self, root: TreeNode, p: int, q: int) -> int: def _lca(r, p, q): """find lca""" if not r: return if r.val == p or r.val == q: return r left = _lca(r.left, p, q) right = _lca(r.right, p, q) if left and right: return r if not left: return right if not right: return left def dfs(r, target): """ we either search target on the left subtree of lca or right subtree of lca to calculate distance """ if not r: return float("-inf") if r.val == target: return 0 return 1 + max(dfs(r.left, target), dfs(r.right, target)) lca = _lca(root, p, q) d1, d2 = dfs(lca, p), dfs(lca, q) return d1+d2
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