LC.P236[二叉树的最近公共祖先]

方法一：DFS+回溯

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        List<TreeNode> a = new ArrayList<>(), b = new ArrayList<>();
        dfs(root, p, a);
        dfs(root, q, b);
        TreeNode ans = null;
        for (int i = 0; i < Math.min(a.size(), b.size()) && a.get(i) == b.get(i); ++i) {
            ans = a.get(i);
        }
        return ans;
    }

    private boolean dfs(TreeNode cur, TreeNode target, List<TreeNode> path) {
        if (cur == null) return false;
        path.add(cur);
        if (cur == target || dfs(cur.left, target, path) || dfs(cur.right, target, path)) {
            return true;
        } else {
            path.remove(path.size() - 1);
            return false;
        }
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：DFS

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q) return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if(left == null && right == null) return null;
        if(left == null) return right;
        if(right == null) return left;
        return root; // left != null and right != null
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$