LeetCode Lowest Common Ancestor of a Binary Tree

Description:

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest
common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

_______3______
/              \
___5__          ___1__
/      \        /      \
6      _2       0       8
/  \
7   4

For example, the lowest common ancestor (LCA) of nodes

5

and

1

is

3

.
Another example is LCA of nodes

5

and

4

is

5

,
since a node can be a descendant of itself according to the LCA definition.

Solution:

LCA的解法有很多，但是这里又是一个没有数据范围的题目……( ▼-▼ )

比较优雅的做法是用后序遍历将两个节点p和q的祖先点都找到，然后找最后一个相同祖先节点就是LCA

注：后序遍历可以用于输出祖先节点

<span style="font-size:18px;">import java.util.*;

public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
LinkedList<TreeNode> n1 = dfs(root, p, new LinkedList<TreeNode>());
LinkedList<TreeNode> n2 = dfs(root, q, new LinkedList<TreeNode>());

TreeNode lca = root;
int n = Math.min(n1.size(), n2.size());
for (int i = 0; i < n; i++) {
if (n1.get(i) == n2.get(i))
lca = n1.get(i);
else
return lca;
}

return lca;
}

public LinkedList<TreeNode> dfs(TreeNode root, TreeNode target,
LinkedList<TreeNode> list) {
if (root == null)
return null;
list.add(root);
if (root == target)
return list;
LinkedList<TreeNode> node;
node = dfs(root.left, target, list);
if (node != null)
return list;
node = dfs(root.right, target, list);
if (node != null)
return list;
list.remove(root);
return null;
}

public static void main(String[] args) {
Solution s = new Solution();

TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);
TreeNode n3 = new TreeNode(3);
TreeNode n4 = new TreeNode(4);
TreeNode n5 = new TreeNode(5);
TreeNode n6 = new TreeNode(6);
n1.left = n2;
n2.left = n4;
n2.right = n5;
n1.right = n3;
n3.right = n6;

System.out.println(s.lowestCommonAncestor(n1, n5, n2));
}
}</span>