Given inorder and postorder traversal of a tree, construct the binary tree.
与Construct Binary Tree from Inorder and Preorder Traversal问题非常类似,唯一区别在于这一次确定root的位置由post traversal来确定,为最后一个元素。
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 if (inorder.length==0 || postorder.length==0 || inorder.length!=postorder.length) { 13 return null; 14 } 15 int len = inorder.length; 16 HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); 17 for (int i=0; i<inorder.length; i++) { 18 map.put(inorder[i], i); 19 } 20 return helper(postorder, 0, len-1, inorder, 0, len-1, map); 21 } 22 23 public TreeNode helper(int[] postorder, int posL, int posR, int[] inorder, int inL, int inR, HashMap<Integer, Integer> map) { 24 if (posL>posR || inL>inR) { 25 return null; 26 } 27 TreeNode root = new TreeNode(postorder[posR]); 28 int index = map.get(postorder[posR]); 29 root.left = helper(postorder, posL, posL+index-1-inL, inorder, inL, index-1, map); 30 root.right = helper(postorder, posL+index-inL, posR-1, inorder, index+1, inR, map); 31 return root; 32 } 33 }