Lowest Common Ancestor of a Binary Tree（二叉树公共祖先）

来源：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree

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.

递归搜索，当root非空时，对其左右两子数分别进行搜索。若搜索结果都为非空，则两个节点分别位于左右两个子树中，LCA为root节点；若其中一个结果为空，则另一个不为空的为LCA；若两个结果都为空，则返回null

 /**
  * 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 || (p == null && q == null)) {
             return null;
         }
         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 && right == null) {
             return left;
         } else if(left == null && right != null) {
             return right;
         } else {
             return null;
         }
     }
 }