Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
根据树的中序遍历和后序遍历,来恢复树,使用递归的思想。
TreeNode getTree(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) { if (inStart >= inEnd) { return null; } int cur = postorder[postEnd - 1]; TreeNode root = new TreeNode(cur); int i = 0; int j = inStart; while(inorder[j] != cur) { j++; i++; } root.left = getTree(inorder, inStart, inStart + i, postorder, postStart, postStart + i); root.right = getTree(inorder, inStart + i + 1, inEnd, postorder, postStart + i, postEnd - 1); return root; } public TreeNode buildTree(int[] inorder, int[] postorder) { if (inorder == null) { return null; } int len = inorder.length; return getTree(inorder, 0, len, postorder, 0, len); }
Construct Binary Tree from Inorder and Postorder Traversal,布布扣,bubuko.com
Construct Binary Tree from Inorder and Postorder Traversal
原文地址:http://blog.csdn.net/u010378705/article/details/28607977