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;
    }
}