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原题链接在这里：http://www.lintcode.com/en/problem/subtree/
You have two every large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree ofT1.
T2 is a subtree of T1 in the following case:
1 3
/ \ /
T1 = 2 3 T2 = 4
/
4
T2 isn‘t a subtree of T1 in the following case:
1 3
/ \ T1 = 2 3 T2 = 4
/
4
A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.
Time Complexity: O(m*n), m是T1的node数, n 是T2的node 数. Space: O(logm). isSame用logn, isSubtree用logm.
AC Java:
1 /** 2 * Definition of TreeNode: 3 * public class TreeNode { 4 * public int val; 5 * public TreeNode left, right; 6 * public TreeNode(int val) { 7 * this.val = val; 8 * this.left = this.right = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 /** 14 * @param T1, T2: The roots of binary tree. 15 * @return: True if T2 is a subtree of T1, or false. 16 */ 17 public boolean isSubtree(TreeNode T1, TreeNode T2) { 18 // write your code here 19 if(T2 == null){ 20 return true; 21 } 22 if(T1 == null){ 23 return false; 24 } 25 return isSame(T1,T2) || isSubtree(T1.left, T2) || isSubtree(T1.right, T2); 26 } 27 28 private boolean isSame(TreeNode T1, TreeNode T2){ 29 if(T1 == null && T2 == null){ 30 return true; 31 } 32 if(T1 == null || T2 == null){ 33 return false; 34 } 35 if(T1.val != T2.val){ 36 return false; 37 } 38 return isSame(T1.left, T2.left) && isSame(T1.right, T2.right); 39 } 40 }
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5188429.html
