Given the root of a binary tree, return the lowest common ancestor of its deepest leaves.
Recall that:
The node of a binary tree is a leaf if and only if it has no children
The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1.
The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A.
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.
Example 2:
Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree, and it's the lca of itself.
Example 3:
Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest leaf node in the tree is 2, the lca of one node is itself.
Constraints:
The number of nodes in the tree will be in the range [1, 1000].
We design a function dfs(root) that returns a tuple (l, d), where l is the deepest common ancestor of node root, and d is the depth of node root. The execution logic of the function dfs(root) is as follows:
If root is null, return the tuple (None, 0);
Otherwise, we recursively call dfs(root.left) and dfs(root.right), obtaining tuples (l, d1) and (r, d2). If d1 > d2, the deepest common ancestor of root is l, and the depth is d1 + 1; if d1 < d2, the deepest common ancestor of root is r, and the depth is d2 + 1; if d1 = d2, the deepest common ancestor of root is root, and the depth is d1 + 1.
In the main function, we call dfs(root) and return the first element of its return value to get the deepest common ancestor node.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.
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# 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 = rightclassSolution:deflcaDeepestLeaves(self,root:Optional[TreeNode])->Optional[TreeNode]:defdfs(root):ifrootisNone:returnNone,0l,d1=dfs(root.left)r,d2=dfs(root.right)ifd1>d2:returnl,d1+1ifd1<d2:returnr,d2+1returnroot,d1+1returndfs(root)[0]
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */typepairstruct{first*TreeNodesecondint}funclcaDeepestLeaves(root*TreeNode)*TreeNode{vardfsfunc(root*TreeNode)pairdfs=func(root*TreeNode)pair{ifroot==nil{returnpair{nil,0}}l,r:=dfs(root.Left),dfs(root.Right)d1,d2:=l.second,r.secondifd1>d2{returnpair{l.first,d1+1}}ifd1<d2{returnpair{r.first,d2+1}}returnpair{root,d1+1}}returndfs(root).first}