[leedcode 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.

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
List<TreeNode> list1;
List<TreeNode> list2;
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

/*      前序遍历二叉树，查找p和q。
一旦找到某一节点，就将其返回，不再遍历其子树。
若某一子树遍历完后都没有找到p或q，则返回null。
每一次递归完成后，比较当前根节点的左右子树是否找到p或q。
若左右子树都找到了（p和q分别在左子树和右子树中），返回当前根节点。
若只有一个子树找到了，另一子树返回为空（p和q都在同一子树中），返回不为空的子树节点（不一定是当前根节点的子树根节点）。
若两个子树都没找到（p和q都不在该子树中），返回左子树（实则返回null）。*/

//注意此时lowestCommonAncestor函数返回值意义已经变了，代表从root起点，是否存在两个节点（之一即可）

/*if(root==null) return null;
if(p==root||q==root) 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?left:right;*/

//第二种方法：使用两个链表保存从根节点到指定节点的路径中遍历到的值，然后转变为求两个链表第一个不同的节点
list1=new ArrayList<TreeNode>();
list2=new ArrayList<TreeNode>();
getPath(root,p,list1);
getPath(root,q,list2);
int i=0;
int j=0;
for(;i<list1.size()&&j<list2.size();i++,j++){
if(list1.get(i)!=list2.get(j)) break;
}
return list1.get(i-1);
}
public boolean getPath(TreeNode root,TreeNode p,List<TreeNode> list){
if(root==null) return false;
list.add(root);
if(root==p) return true;
boolean isExist=false;
isExist=getPath(root.left,p,list);
if(!isExist){
isExist=getPath(root.right,p,list);

}
if(!isExist){
list.remove(list.size()-1);
return false;
}else return true;

}
}