面試題之樹中兩個節點的最低公共祖先節點
1.樹為二叉搜尋樹,二叉搜尋樹特點:值:右>根>左
①思路
從根節點開始遍歷
a:若該節點值比所給兩個節點均大,則公共祖先節點必在其左子樹上,遍歷其左子樹。
b:若該節點值比所給兩個節點均小,則公共祖先節點必在其右子樹上,遍歷其右子樹。
c:直到找到一個節點,位於兩節點的中間。
Given a binary search tree(BST),find the lowest common ancestor(LCA) of two given nodes in the BST.
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)."
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6.Another example is LCA of nodes 2 and 4 is 2,since a node can be a descendant of itself according to the LCA definition.
②程式碼
public class BanarySearchTreeMain { public static void main(String[] args) { // TODO Auto-generated method stub TreeNode root = initTree(); TreeNode result = lowestCommonAncestor(root, new TreeNode(2), new TreeNode(4)); System.out.println(result.val); } /* * 初始化二叉搜尋樹,其特點:右>根>左; */ private static TreeNode initTree(){ TreeNode root = new TreeNode(6); TreeNode node0 = new TreeNode(0); TreeNode node2 = new TreeNode(2); TreeNode node3 = new TreeNode(3); TreeNode node4 = new TreeNode(4); TreeNode node5 = new TreeNode(5); TreeNode node7 = new TreeNode(7); TreeNode node8 = new TreeNode(8); TreeNode node9 = new TreeNode(9); root.left = node2; root.right = node8; node2.left = node0; node2.right = node4; node8.left = node7; node8.right = node9; node4.left = node3; node4.right = node5; return root; } /* * 二叉搜尋樹最低公共父節點 */ private static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root==null||root==p||root==q) { return root; } if(root.val>p.val&&root.val>q.val) { return lowestCommonAncestor(root.left, p, q); } if(root.val<p.val&&root.val<q.val) { return lowestCommonAncestor(root.right, p, q); }else { return root; } } } class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } }
2.樹為普通二叉樹
①思路
a:遍歷左子樹,若與兩節點中某值相等或左右均不為null,則返回根節點,否則,繼續遍歷;
b:遍歷右子樹,若與兩節點中某值相等或左右均不為null,則返回根節點,否則,繼續遍歷;
c:返回節點;
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.
②程式碼
public class CommonTreeMain {
public static void main(String[] args) {
TreeNode root = initTree();
TreeNode result = lowestCommonAncestor(root, new TreeNode(5), new TreeNode(4));
System.out.println(result.val);
}
private static TreeNode initTree(){
TreeNode root = new TreeNode(3);
TreeNode node0 = new TreeNode(0);
TreeNode node1 = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node4 = new TreeNode(4);
TreeNode node5 = new TreeNode(5);
TreeNode node6 = new TreeNode(6);
TreeNode node7 = new TreeNode(7);
TreeNode node8 = new TreeNode(8);
root.left = node5;
root.right = node1;
node5.left = node6;
node5.right = node2;
node1.left = node0;
node1.right = node8;
node2.left = node7;
node2.right = node4;
return root;
}
/*
* 查詢最低公共祖先節點
*/
private static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null||root.val==p.val||root.val==q.val)
return root;
TreeNode left = lowestCommonAncestor(root.left,p,q);
TreeNode right = lowestCommonAncestor(root.right,p,q);
if (left == null)
return right;
if (right == null)
return left;
return root;
}
}
③迭代演算法:
需要我們儲存下由root根節點到p和q節點的路徑,並且將路徑存入list中,則問題轉化為求兩個list集合的最後一個共同元素。
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null || p==null || q==null)
return null;
List<TreeNode> pathp = new ArrayList<>();
List<TreeNode> pathq = new ArrayList<>();
pathp.add(root);
pathq.add(root);
getPath(root, p, pathp);
getPath(root, q, pathq);
TreeNode lca = null;
for(int i=0; i<pathp.size() && i<pathq.size(); i++) {
if(pathp.get(i) == pathq.get(i))
lca = pathp.get(i);
else
break;
}
return lca;
}
private boolean getPath(TreeNode root, TreeNode n, List<TreeNode> path) {
if(root==n)
return true;
if(root.left!=null) {
path.add(root.left);
if(getPath(root.left, n, path))
return true;
path.remove(path.size()-1);
}
if(root.right!=null) {
path.add(root.right);
if(getPath(root.right, n, path))
return true;
path.remove(path.size()-1);
}
return false;
}
轉載:http://blog.csdn.net/qq_25827845/article/details/74612786