LeetCode: Construct Binary Tree from Inorder and Postorder Traversal 解题报告

Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

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Tree Array Depth-first Search

 

SOLUTION 1:

使用递归的思想，先找到根节点（它就是post order最后一个），然后再在inorder中找到它，以确定左子树的node个数。

然后分别确定左子树右子树的左右边界

例子：

{4， 5， 2， 7， 8， 1， 3}这树的

inorder: 7 5 8 | 4 | 1 2 3

post: 7 8 5 | 1 3 2 | 4

以上我们可以看到左右子树的划分关系。

 /**
  * Definition for binary tree
  * 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) {
             return null;
         }

         return dfs(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
     }

     public TreeNode dfs(int[] inorder, int[] postorder, int inL, int inR, int postL, int postR) {
         if (inL > inR) {
             return null;
         }

         // create the root node.
         TreeNode root = new TreeNode(postorder[postR]);

         // find the position of the root node in the inorder traversal.
         int pos = 0;
         for (; pos <= inR; pos++) {
             if (inorder[pos] == postorder[postR]) {
                 break;
             }
         }

         int leftNum = pos - inL;

         root.left = dfs(inorder, postorder, inL, pos - 1, postL, postL + leftNum - 1);
         root.right = dfs(inorder, postorder, pos + 1, inR, postL + leftNum, postR - 1);

         return root;
     }
 }

代码： https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/BuildTree2.java