construct the binary tree.

一、题目

题目描述Given inorder and postorder traversal of a tree, construct the binary tree.Note:
You may assume that duplicates do not exist in the tree.

二、分析

中序:DBEAC
后序:DEBCA

通过结合图片和中序,后序序列可以很清楚地看到A是这棵树的根节点,并且有左右子树,左子树中有B,D,E,而右子树中只有一个C。通过观察后序序列可以发现根节点位于此序列的最右端,而根节点在中序序列中的左边是它的左子树,右边是它的右子树。根据这样的特性得出以下代码:


public class ConstructBinaryTree {public class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}/*** * @param inorder   中序序列* @param postorder  后序序列* @return*/public TreeNode buildTree(int[] inorder, int[] postorder) {if (inorder == null || postorder == null || inorder.length != postorder.length)return null;int starti = 0;int endi = inorder.length - 1;int startp = 0;int endp = postorder.length - 1;TreeNode root = build(inorder, postorder, starti, endi, startp, endp);return root;}public TreeNode build(int[] inorder, int[] postorder, int starti, int endi, int startp, int endp) {if (starti > endi || startp > endp)return null;TreeNode root = new TreeNode(postorder[endp]);int j = 0;for (int i = starti; i <= endi; i++) {if (root.val == inorder[i])j = i;//记录root节点在中序遍历中的位置}int len = j - starti;//左的长度//递归分别对root节点的左子树和右子树进行buildroot.left = build(inorder, postorder, starti, j - 1, startp, startp + len - 1);root.right = build(inorder, postorder, j + 1, endi, startp + len, endp - 1);return root;}}

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construct the binary tree.

一、题目

二、分析

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