Lowest Common Ancestor of a Binary Tree

Posted on |

问题描述（难度中等）

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

img

Example 1:

1
2
3
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

1
2
3
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

  • All of the nodes’ values will be unique.
  • p and q are different and both values will exist in the binary tree.

方法一：递归实现

递归的方式，左右子树递归，返回一个符合条件的节点。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
package P236;
import CommonUtils.TreeNodeUtils;
import P104.TreeNode;

/**
 * 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.val==p.val || root.val==q.val){
            return root;
        }
        TreeNode left=lowestCommonAncestor(root.left,p,q);
        TreeNode right=lowestCommonAncestor(root.right,p,q);
        if (left!=null && right!=null) {
            return root;
        }
        return left!=null?left:right;
    }

    public static void main(String[] args) {
        int[] ints={3,5,1,6,2,0,8,7,4};
        TreeNode treeNode=TreeNodeUtils.buildTreeNodeUsingArray(ints);
        new Solution().lowestCommonAncestor(treeNode,treeNode.left.left,treeNode.right);
    }
}

算法参考左序遍历：

cmd-markdown-logo