Given inorder and postorder traversal of a tree, construct the binary tree.
题目大意:给定一个二叉树的中序和后续序列,构建出这个二叉树。
解题思路:首先后序序列的最后一个是根节点,然后在中序序列中找到这个节点,中序序列中这个节点左边的是根节点的左子树,右边的是右子树,由此递归构建出完整的树。
Talk is cheap:
public TreeNode buildTree(int[] inorder, int[] postorder) { if (inorder == null || postorder == null) { return null; } int inLen = inorder.length; int postLen = postorder.length; if ((inLen == 0 && postLen == 0) || inLen != postLen) { return null; } TreeNode root = new TreeNode(postorder[postLen - 1]); if (inLen == 1) { return root; } int pos = 0; for (int i = 0; i < inLen; i++) { if (inorder[i] == postorder[postLen - 1]) { pos = i; break; } } int[] inLeft = Arrays.copyOfRange(inorder, 0, pos); int[] inRight = Arrays.copyOfRange(inorder, pos + 1, inLen); int[] postLeft = Arrays.copyOfRange(postorder, 0, pos); int[] postRight = Arrays.copyOfRange(postorder, pos, postLen - 1); root.left = buildTree(inLeft, postLeft); root.right = buildTree(inRight, postRight); return root; }