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106. Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { public TreeNode buildTree(int[] inorder, int[] postorder) { return buildSubTree(0, postorder.length -1, postorder, inorder, 0, postorder.length-1); } public TreeNode buildSubTree(int postLeft, int postRight, int[] postorder, int[] inorder, int inLeft, int inRight){ if(postRight < postLeft || inLeft > inRight) return null; int mid = inLeft; TreeNode root = new TreeNode(postorder[postRight]); while(mid <= inRight){ if(postorder[postRight] == inorder[mid]){ root.left = buildSubTree(postLeft, postRight+mid-inRight -1 , postorder, inorder, inLeft, mid-1); root.right = buildSubTree(postRight+mid-inRight, postRight -1, postorder, inorder, mid+1, inRight); } mid++; } return root; }}
106. Construct Binary Tree from Inorder and Postorder Traversal
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