leetcode solution 106. Construct Binary Tree from Inorder and Postorder Traversal
z
 2018-03-07 16:06:26  

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

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

For example, given

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2
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

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2
3
4
5
  3
 / \
9  20
  /  \
 15   7
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder.length == 0 || postorder.length == 0)
            return null;
        return build(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    public TreeNode build(int[] inorder, int instart, int inend, int[] postorder, int poststart, int postend){
        if(instart>inend || poststart > postend)
            return null;
        TreeNode root = new TreeNode(postorder[postend]);
        int cur = -1;
        for(int i=instart; i<= inend; i++){
            if(root.val == inorder[i]){
                cur = i;
                break;
            }
        }
        root.left = build(inorder, instart, cur-1, postorder, poststart, postend-(inend - cur)-1);
        root.right = build(inorder, cur+1, inend, postorder, postend-(inend - cur), postend-1);
        return root;
    }
}
  • 注意start end 的取值