236. Lowest Common Ancestor of a Binary Tree（最低公共祖先，难理解）

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 5and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

更新于20180513

若pq都在某个节点的左边，就到左子树中查找，如果都在右边 就到右子树种查找。

要是pq不在同一边，那就表示已经找到第一个公共祖先。

 class Solution {
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
         if(!cover(root,p)||!cover(root,q)) return null;
         return LCAhelp(root,p,q);
     }
     private  TreeNode LCAhelp(TreeNode root,TreeNode p,TreeNode q ){
         if(root==null) return null;
         if(root==p||root==q) return root;
         boolean q_is_on_left = cover(root.left,q);
         boolean p_is_on_left = cover(root.left,p);

         //分立两边
         if(q_is_on_left!=p_is_on_left) return root;

         //在一边
         else{
             if(q_is_on_left)
                 return LCAhelp(root.left,p,q);
             else
                 return LCAhelp(root.right,p,q);
         }
     }
     private boolean cover(TreeNode root,TreeNode p){
         //检查p是不是root的孙子

         if(root==null) return false;
         if(root==p) return true;
         return cover(root.left,p)||cover(root.right,p);
     }
 }

解题思路

• Divide & Conquer 的思路
• 如果root为空，则返回空
• 如果root等于其中某个node，则返回root
• 如果上述两种情况都不满足，则divide，左右子树分别调用该方法
• Divide & Conquer中治这一步要考虑清楚，本题三种情况
• 如果left和right都有结果返回，说明root是最小公共祖先
• 如果只有left有返回值，说明left的返回值是最小公共祖先
• 如果只有�right有返回值，说明�right的返回值是最小公共祖先

 class Solution {
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
         if(root ==null || p==root||q==root) return root;
         TreeNode lp = lowestCommonAncestor(root.left,p,q);
         TreeNode rp = lowestCommonAncestor(root.right,p,q);
         if(lp!=null&&rp!=null) return root;
         if(lp==null&&rp!=null) return rp;
         if(lp!=null&&rp==null) return lp;
         return null;
     }
 }