leetcode--Construct Binary Tree from Preorder and Inorder Traversal
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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
分类:二叉树
题意:由前序遍历,中序遍历生成二叉树
解法1:递归。将前序遍历和中序遍历分成两部分。对于前序遍历而言,数组第一个元素为根节点。在中序遍历中找到这个根节点,其左右两边则分别是左右子树的中序遍历。
计算左子树长度len(根据根节点index求出),然后对应于前序遍历,从根节点(数组第一个节点)开始,len长度之内,同样是左子树的前序遍历,剩下的是右子树的前序遍历。
这样我们又得到两个中序遍历和前序遍历,递归求解即可。
/** * 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[] preorder, int[] inorder) {return hepler(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1); }public TreeNode hepler(int[] preorder,int pstart,int pend,int[] inorder,int istart,int iend){ if(pstart>pend) return null;if(pstart==pend) return new TreeNode(preorder[pstart]);int f = preorder[pstart];TreeNode t = new TreeNode(f);int mid = istart;for(;mid<=iend;mid++){if(f==inorder[mid]) break;}int len = mid-istart;t.left = hepler(preorder, pstart+1, pstart+len, inorder, istart, mid-1);t.right = hepler(preorder, pstart+len+1, pend, inorder, mid+1, iend);return t;}}
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