Lowest Common Ancestor of a Binary Tree——Leetcode

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.

题目大意：求二叉树的两个节点的最近祖先。

解题思路：递归的来写，

设两个变量left和right，如果在root节点的左右孩子节点分别发现节点p和q，

那么假设left＝q，right＝p

那么root就是最近祖先；

如果left＝null，right＝q或p，那么right就是最近祖先；

如果right＝null，left＝q或p，那么left就是最近祖先。

　　public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null||root==q||root==p){
            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;
    }
    

 也看到有人用map记录父节点信息，然后利用map找出最近公共祖先，也不错。

代码如下：

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    HashMap<TreeNode,TreeNode> m = new HashMap<TreeNode,TreeNode>();
    Queue<TreeNode> queue = new LinkedList<TreeNode>();
    queue.offer(root);
    while(queue.peek()!=null){
        TreeNode t = queue.poll();
        if(t.left!=null){
            m.put(t.left,t);
            queue.offer(t.left);
        }
        if(t.right!=null){
            m.put(t.right,t);
            queue.offer(t.right);
        }
    }
    Set<TreeNode> l = new HashSet<TreeNode>();
    TreeNode pp = p;
    while(pp!=root){
        l.add(pp);
        pp = m.get(pp);
    }
    l.add(root);
    TreeNode qq = q;
    while(!l.contains(qq)){
        qq = m.get(qq);
    }
    return qq;
}