You are given a binary search tree root and two nodes p and q. Find the lowest common ancestor (LCA) of nodes p and q. The lowest common ancestor is defined as the lowest node in root that has both p and q as descendants (where we allow a node to be a descendant of itself). Select the optimal solution in terms of time complexity. If time complexities are same for multiple solutions, select the one with optimal space complexity. Include recursion stack in the space complexity. function lca(root, p, q): if not root: return None left = lca(root.left, p, q) right = lca(root.right, p, q) if left and right: return root return left if left else right function lca(root, p, q): if p.val < root.val and q.val < root.val: return lca(root.left, p, q) elif p.val > root.val and q.val > root.val: return lca(root.right, p, q) else: return root function lca(root, p, q): while root: if p.val < root.val and q.val < root.val: root = root.left elif p.val > root.val and q.val > root.val: root = root.right else: return root
You are given a binary search tree root and two nodes p and q. Find the lowest common ancestor (LCA) of nodes p and q. The lowest common ancestor is defined as the lowest node in root that has both p and q as descendants (where we allow a node to be a descendant of itself). Select the optimal solution in terms of time complexity. If time complexities are same for multiple solutions, select the one with optimal space complexity. Include recursion stack in the space complexity. function lca(root, p, q): if not root: return None left = lca(root.left, p, q) right = lca(root.right, p, q) if left and right: return root return left if left else right function lca(root, p, q): if p.val < root.val and q.val < root.val: return lca(root.left, p, q) elif p.val > root.val and q.val > root.val: return lca(root.right, p, q) else: return root function lca(root, p, q): while root: if p.val < root.val and q.val < root.val: root = root.left elif p.val > root.val and q.val > root.val: root = root.right else: return root
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You are given a binary search tree root and two nodes p and q. Find the lowest common ancestor (LCA) of nodes p and q. The lowest common ancestor is defined as the lowest node in root that has both p and q as descendants (where we allow a node to be a descendant of itself). Select the optimal solution in terms of time complexity. If time complexities are same for multiple solutions, select the one with optimal space complexity. Include recursion stack in the space complexity.
function lca(root, p, q):
if not root:
return None
left = lca(root.left, p, q)
right = lca(root.right, p, q)
if left and right:
return root
return left if left else right
if not root:
return None
left = lca(root.left, p, q)
right = lca(root.right, p, q)
if left and right:
return root
return left if left else right
function lca(root, p, q):
if p.val < root.val and q.val < root.val:
return lca(root.left, p, q)
elif p.val > root.val and q.val > root.val:
return lca(root.right, p, q)
else:
return root
if p.val < root.val and q.val < root.val:
return lca(root.left, p, q)
elif p.val > root.val and q.val > root.val:
return lca(root.right, p, q)
else:
return root
function lca(root, p, q):
while root:
if p.val < root.val and q.val < root.val:
root = root.left
elif p.val > root.val and q.val > root.val:
root = root.right
else:
return root
while root:
if p.val < root.val and q.val < root.val:
root = root.left
elif p.val > root.val and q.val > root.val:
root = root.right
else:
return root
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