Note:
You may assume that duplicates do not exist in the tree.
用后序遍历序列来确定根节点的位置,然后在中序遍历中确定左右子树,递归完成树的构造。代码如下:
/** * 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) { if(inorder == null || postorder == null || inorder.length == 0 || postorder.length == 0) return null; TreeNode root = new TreeNode(postorder[postorder.length - 1]); int i; for(i = 0; i < inorder.length; i++) { if(inorder[i] == postorder[postorder.length-1]) break; } int[] lp = Arrays.copyOfRange(postorder, 0, i); int[] li = Arrays.copyOfRange(inorder, 0, i); int[] rp = Arrays.copyOfRange(postorder, i, postorder.length - 1); int[] ri = Arrays.copyOfRange(inorder, i + 1, inorder.length); root.left = buildTree(li, lp); root.right = buildTree(ri, rp); return root; } }