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Given an integer array with no duplicates. A max tree building on this array is defined as follow:
Given [2, 5, 6, 0, 3, 1], the max tree is
6
/ \
5 3
/ / \
2 0 1
O(n) time complexity
Analysis:
Recursion: use recursion method, in the worst case, the complexity is O(n^2).
Linear method: Refer to the analysis of some other people: http://www.meetqun.com/thread-3335-1-1.html
这个题Leetcode上没有，其实这种树叫做笛卡树（ Cartesian tree）。直接递归建树的话复杂度最差会退化到O(n^2)。经典建树方法，用到的是单调堆栈。我们堆栈里存放的树，只有左子树，没有有子树，且根节点最大。
（1） 如果新来一个数，比堆栈顶的树根的数小，则把这个数作为一个单独的节点压入堆栈。
（2） 否则，不断从堆栈里弹出树，新弹出的树以旧弹出的树为右子树，连接起来，直到目前堆栈顶的树根的数大于新来的数。然后，弹出的那些数，已经形成了一个新的树，这个树作为新节点的左子树，把这个新树压入堆栈。
这样的堆栈是单调的，越靠近堆栈顶的数越小。
最后还要按照（2）的方法，把所有树弹出来，每个旧树作为新树的右子树。
Solution:
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 A: Given an integer array with no duplicates. 15 * @return: The root of max tree. 16 */ 17 public TreeNode maxTree(int[] A) { 18 if (A.length==0) return null; 19 20 Stack<TreeNode> nodeStack = new Stack<TreeNode>(); 21 nodeStack.push(new TreeNode(A[0])); 22 for (int i=1;i<A.length;i++) 23 if (A[i]<=nodeStack.peek().val){ 24 TreeNode node = new TreeNode(A[i]); 25 nodeStack.push(node); 26 } else { 27 TreeNode n1 = nodeStack.pop(); 28 while (!nodeStack.isEmpty() && nodeStack.peek().val < A[i]){ 29 TreeNode n2 = nodeStack.pop(); 30 n2.right = n1; 31 n1 = n2; 32 } 33 TreeNode node = new TreeNode(A[i]); 34 node.left = n1; 35 nodeStack.push(node); 36 } 37 38 39 TreeNode root = nodeStack.pop(); 40 while (!nodeStack.isEmpty()){ 41 nodeStack.peek().right = root; 42 root = nodeStack.pop(); 43 } 44 45 return root; 46 47 48 49 50 51 } 52 }
Solution 2 (recursion):
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 A: Given an integer array with no duplicates. 15 * @return: The root of max tree. 16 */ 17 public TreeNode maxTree(int[] A) { 18 if (A.length==0) return null; 19 20 TreeNode root = new TreeNode(A[0]); 21 for (int i=1;i<A.length;i++) 22 if (A[i]>root.val){ 23 TreeNode node = new TreeNode(A[i]); 24 node.left = root; 25 root = node; 26 } else insertNode(root,null,A[i]); 27 28 return root; 29 } 30 31 public void insertNode(TreeNode cur, TreeNode pre, int val){ 32 if (cur==null){ 33 TreeNode node = new TreeNode(val); 34 pre.right = node; 35 return; 36 } 37 38 if (cur.val<val){ 39 TreeNode node = new TreeNode(val); 40 pre.right = node; 41 node.left = cur; 42 return; 43 } else 44 insertNode(cur.right,cur,val); 45 } 46 47 48 }
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原文地址：http://www.cnblogs.com/lishiblog/p/4187936.html
