Lowest Common Ancestor of a Binary Tree

Posted 2020-06-21 YuriFLAG

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.

可以采用Divide&conquer。

思路：1.将问题变为寻找以左子树为根节点的LCA 和 以右子树为根节点的LAC。

　　　2.如果以左子树为根节点的LCA 和 以右子树为根节点的LAC都存在表明，该节点的左孩子和右孩子是p 和 q ，因此LCA为左右子树的根节点。

　　　3.如果以左子树为根节点的LCA为空，以右子树为根节点的LAC不为空，则返回右子树为根节点的LCA。

　　　4.如果以右子树为根节点的LCA为空，以左子树为根节点的LAC不为空，则返回左子树为根节点的LCA。

　　　5.如果以左子树为根节点的LCA 和 以右子树为根节点的LAC都不存在返回null.

　　　　

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
12         if (root == null || p == null || q == null) {
13             return null;
14         }
15         if (p == root || q == root) {
16             return root;
17         }
18         //Divide
19         TreeNode left = lowestCommonAncestor(root.left, p, q);
20         TreeNode right = lowestCommonAncestor(root.right, p, q);
21         //Conquer
22         if (left != null && right != null) {
23             return root;
24         }
25         if (left == null && right != null) {
26             return right;
27         }
28         if (left != null && right == null) {
29             return left;
30         }
31         return null;
32     }
33 }