【LeetCode】236 Lowest Common Ancestor of a Binary Tree

Lowest Common Ancestor of a Binary Tree
Total Accepted: 1207 Total Submissions: 4143 My Submissions Question Solution 

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.

【解题思路】

二叉树，需要左右孩子都搜索，典型DFS。但是防止做孩子搜索了，右孩子又重复搜索，需要判断返回结果。

Java AC

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        List list1 = new ArrayList();
        List list2 = new ArrayList();
        getPath(root, p, list1);
        getPath(root, q, list2);
        int size1 = list1.size();
        int size2 = list2.size();
        int len = size1 < size2 ? size1 : size2;
        int k = 0;
        while(k < len){
            TreeNode node1 = list1.get(k);
            TreeNode node2 = list2.get(k);
            if(node1 != node2){
                break;
            }
            k++;
        }
        return list1.get(k - 1);
    }

    private boolean getPath(TreeNode root, TreeNode d, List list) {
        if (root == d){
            list.add(root);
            return true;
        }
        if (root != null){
            list.add(root);
            if (getPath(root.left, d, list) || getPath(root.right, d, list)){
                return true;
            }
            list.remove(list.size()-1);
        }
        return false;
    }
}