236. 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.

这个题目是找二叉树两个节点的公共祖先，之前有一个题目是二叉搜索树。相比，二叉搜索树显然更容易，因为节点大小是有序的。

需要多考虑的一个问题就是 如果给定的两个节点中有一个恰就是最近公共祖先，那么在搜寻另一个分支时是搜索不到结果的也就是返回null；

如果某一支返回null且同级的另一个含有目标值，当两侧有两个目标值在某一级相遇时说明这时的根节点是公共祖先

返回值从最底层不断向上走 直到两侧都相遇找到公共祖先 

           __3______       /              \    ___5__          ___1__   /      \        /      \   6      _2      0       8         /  \         3  4

public class Solution {    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {        if(root==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 ;        return left==null?right:left;    }    }