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leetcode 94. Binary Tree Inorder Traversal

时间:2016-08-13 19:30:27      阅读:172      评论:0      收藏:0      [点我收藏+]

 

 

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree [1,null,2,3],

   1
         2
    /
   3

 

return [1,3,2].

 

非递归方法写二叉树的中序遍历

 

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        List<Integer> inorder = new ArrayList<Integer>();
        
        if (root  == null) {
            return inorder;
        }
        
        TreeNode curt = root;
        while (curt != null || !stack.empty()) {
            while (curt != null) {
                stack.push(curt);
                curt = curt.left;
            }
            
            curt = stack.pop();
            inorder.add(curt.val);
            curt = curt.right;
        }
        return inorder;
    }
}

 

leetcode 94. Binary Tree Inorder Traversal

原文:http://www.cnblogs.com/iwangzheng/p/5768525.html

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