leetcode 105 Construct Binary Tree from Preorder and Inorder Traversal ----- java

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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.

 

給出前序走訪和中序遍曆，然後求這棵樹。

 

很有規律。遞迴就可以實現。

/** * 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) {        int len = preorder.length;        if( len == 0)            return null;        return helper(preorder,0,len-1,inorder,0,len-1);            }    public TreeNode helper( int[] preorder,int pre_start,int pre_end,int[] inorder,int in_start,int in_end){        if( pre_start > pre_end || in_start > in_end )            return null;        TreeNode node = new TreeNode(preorder[pre_start]);        int size = 0;        for( int i = in_start;i<=in_end && inorder[i] != preorder[pre_start];i++,size++)            ;        node.left = helper(preorder,pre_start+1,pre_start+size,inorder,in_start,in_start+size-1);        node.right = helper(preorder,pre_start+size+1,pre_end,inorder,in_start+size+1,in_end);        return node;    }}

速度不算快，速度最快的答案也進行了參考。

 

並不是演算法有多好，只是他在

for( int i = in_start;i<=in_end && inorder[i] != preorder[pre_start];i++,size++)            ;

這裡遍曆的時候選擇了從後向前遍曆，由於測試資料的特殊性，導致了其答案的快速性。

 

leetcode 105 Construct Binary Tree from Preorder and Inorder Traversal ----- java