LeetCode Lowest Common Ancestor of a Binary Tree

原题链接在这里：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

题目：

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.

题解：

是Lowest Common Ancestor of a Binary Search Tree进阶题目。

无法比较大小，但是可以看p,q是不是在root的两边，若在两边，left 和 right 同时不失null, 则返回root.

若都在一边，比如left, 就在left边继续。

Note:1. 若有root == p || root == q时，需比较原来的点而不单单是val, 这里可以有重复的值，在能比较整体点时就不比较val.

2. 递归终止条件这里有两个，一个是root == null, 一个是root等于p或者q, 这两个终止条件缺一不可。

Time Complexity: O(n), 每个点没有traverse超过两遍. Space: O(logn), 是树的高度。

AC Java:

 /**
  * 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) {
         if(root == null){
             return root;
         }
         if(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 root;
         }else if(left != null){
             return left;
         }else{
             return right;
         }
     }
 }

跟上Smallest Subtree with all the Deepest Nodes.

类似Smallest Common Region.