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Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
后序遍历的最后一个元素就是根元素,由于没有重复,就在中序遍历的数组中查找根元素,这样就分成的两段,然后递归。
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public TreeNode buildTree(int[] inorder, int[] postorder) { 12 int length=inorder.length; 13 if (length==0) return null; 14 int rootnumber=postorder[length-1]; 15 TreeNode root = new TreeNode(rootnumber); 16 17 if (length==1) return root; 18 int indexRoot = 0; 19 for (int i = 0; i < length; i++) { 20 if (inorder[i]==rootnumber){ 21 indexRoot=i; 22 break; 23 } 24 } 25 int[] left=new int[indexRoot]; 26 int[] leftPost=new int[indexRoot]; 27 int[] right=new int[length-indexRoot-1]; 28 int[] rightPost=new int[length-indexRoot-1]; 29 for (int i = 0; i < indexRoot; i++) { 30 left[i]=inorder[i]; 31 leftPost[i]=postorder[i]; 32 } 33 for (int i = indexRoot+1; i <length ; i++) { 34 right[i-1-indexRoot]=inorder[i]; 35 rightPost[i-1-indexRoot]=postorder[i-1]; 36 } 37 root.left=buildTree(left,leftPost); 38 root.right=buildTree(right,rightPost); 39 return root; 40 } 41 }
LeetCode Construct Binary Tree from Inorder and Postorder Traversal
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原文地址:http://www.cnblogs.com/birdhack/p/4088508.html