Leetcode-Construct Binary Tree from inorder and postorder travesal
Given inorder and postorder traversal of a tree, construct the binary tree.
Solution:
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) 13 return null; 14 15 int len = inorder.length; 16 TreeNode root = buildTreeRecur(inorder,postorder,0,len-1,0,len-1); 17 return root; 18 } 19 20 //Build tree for current list, i.e., inorder[inHead] to inorder[inEnd]. 21 public TreeNode buildTreeRecur(int[] inorder, int[] postorder, int inHead, int inEnd, int postHead, int postEnd){ 22 if (inHead==inEnd){ 23 TreeNode root = new TreeNode(inorder[inHead]); 24 return root; 25 } 26 27 int curRoot = postorder[postEnd]; 28 int index = -1; 29 for (int i=inHead;i<=inEnd;i++) 30 if (inorder[i]==curRoot){ 31 index = i; 32 break; 33 } 34 int leftNodeNum = index-inHead; 35 36 37 int leftInHead = inHead; 38 int leftInEnd = inHead+leftNodeNum-1; 39 int rightInHead = index+1; 40 int rightInEnd = inEnd; 41 42 43 int leftPostHead = postHead; 44 int leftPostEnd = postHead+leftNodeNum-1; 45 int rightPostHead = leftPostEnd+1; 46 int rightPostEnd = postEnd-1; 47 48 TreeNode root = new TreeNode(curRoot); 49 TreeNode leftChild = null; 50 if (leftInEnd>=inHead){ 51 leftChild = buildTreeRecur(inorder,postorder,leftInHead,leftInEnd,leftPostHead,leftPostEnd); 52 root.left = leftChild; 53 } 54 55 TreeNode rightChild = null; 56 if (rightInHead<=inEnd){ 57 rightChild = buildTreeRecur(inorder,postorder,rightInHead,rightInEnd,rightPostHead,rightPostEnd); 58 root.right = rightChild; 59 } 60 61 return root; 62 } 63 }
We need to be very carefull about how to count the start and end of the left sub-tree and the right-sub tree. Especially detecting the case that some sub-tree is void.
A better way is to calculate the number of nodes in left and right tree first, then find out the range, like this:
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) 13 return null; 14 15 int len = inorder.length; 16 TreeNode root = buildTreeRecur(inorder,postorder,0,len-1,0,len-1); 17 return root; 18 } 19 20 //Build tree for current list, i.e., inorder[inHead] to inorder[inEnd]. 21 public TreeNode buildTreeRecur(int[] inorder, int[] postorder, int inHead, int inEnd, int postHead, int postEnd){ 22 if (inHead==inEnd){ 23 TreeNode root = new TreeNode(inorder[inHead]); 24 return root; 25 } 26 27 int curRoot = postorder[postEnd]; 28 TreeNode root = new TreeNode(curRoot); 29 TreeNode leftChild = null; 30 TreeNode rightChild = null; 31 32 int index = -1; 33 for (int i=inHead;i<=inEnd;i++) 34 if (inorder[i]==curRoot){ 35 index = i; 36 break; 37 } 38 int leftNodeNum = index-inHead; 39 int rightNodeNum = inEnd-index; 40 41 if (leftNodeNum>0){ 42 int leftInHead = inHead; 43 int leftInEnd = inHead+leftNodeNum-1; 44 int leftPostHead = postHead; 45 int leftPostEnd = postHead+leftNodeNum-1; 46 leftChild = buildTreeRecur(inorder,postorder,leftInHead,leftInEnd,leftPostHead,leftPostEnd); 47 root.left = leftChild; 48 } 49 50 51 if (rightNodeNum>0){ 52 int rightInHead = index+1; 53 int rightInEnd = inEnd; 54 int rightPostHead = postEnd-rightNodeNum; 55 int rightPostEnd = postEnd-1; 56 rightChild = buildTreeRecur(inorder,postorder,rightInHead,rightInEnd,rightPostHead,rightPostEnd); 57 root.right = rightChild; 58 } 59 60 return root; 61 } 62 }
