Lowest Common Ancestor of Two Nodes in a Binary Tree

本文轉載自查看原文 2014-12-04 08:16 2424 經驗積累/ 面試題/ 知識積累

Reference：

http://blog.csdn.net/v_july_v/article/details/18312089

http://leetcode.com/2011/07/lowest-common-ancestor-of-a-binary-tree-part-ii.html

(1) Is the tree a BST or not?

BST的話，我們就能按照BST的定義思考了。當前遍歷的node如果比我們要找的兩個點都大，說明那兩個點都在當前node的左邊，所以這兩個node的祖先肯定在當前node的左邊，所以往左邊找。反之，往右邊找。如果這兩個點一個大於當前node一個小於當前node，說明當前node就是LCA。如果這兩個點一個是另一個的祖先，那么這個點就是LCA。

代碼如下：

1      public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
2          if(p.val < root.val && q.val < root.val)
3              return lowestCommonAncestor(root.left, p, q);
4          else if(p.val > root.val && q.val > root.val)
5              return lowestCommonAncestor(root.right, p, q);
6          else
7              return root;

8 }

（2）那如果不是BST呢？一般Binary Tree.

a. Tree結構中沒有parent域。

遞歸找。

代碼如下：

1 // Worst case O(n)
2 Node getLCA(Node root, Node node1, Node node2){
3          if(root == null)
4              return null
5          if(root == node1 || root == node2)
6              return root;
7
8         Node left = getLCA(root.left, node1, node2);
9         Node right = getLCA(root.right, node1, node2);
10
11          if(left != null && right != null)
12              return root;
13          else if(left != null)
14              return left;
15          else if(right != null)
16              return right;
17          else
18              return null;
19     }

b. 如果有parent域。

轉化為兩個linkedlist的交點問題。

代碼如下：

1 int getHeight(Node p){
2          int height = 0;
3          while(p!= null){
4             height++;
5             p = p.parent;
6         }
7          return height;
8     }
9
10      void swap( int a, int b){
11         a = a + b;
12         b = a - b;
13         a = a - b;
14     }
15
16     Node getLCA(Node p, Node q){
17          int h1 = getHeight(p);
18          int h2 = getHeight(q);
19
20          // q is always deeper than p
21          if(h1 > h2){
22             swap(h1, h2);
23             swap(p, q);
24         }
25
26          int diff = h2 - h1;
27
28          for( int i = 0; i < diff; i++)
29             q = q.parent;
30
31          while(p!= null && q!= null){
32              // common node
33              if(p == q)
34                  return p;
35             p = p.parent;
36             q = q.parent;
37         }
38
39          return NULL; // p and q are not in the same tree
40     }

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