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;
}
}