LeetCode: Binary Tree Postorder Traversal

最新推荐文章于 2022-07-02 16:50:24 发布

dbu50120

最新推荐文章于 2022-07-02 16:50:24 发布

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原文链接：http://www.cnblogs.com/yeek/p/3486150.html

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本文介绍了一种不使用递归的二叉树后序遍历算法，通过栈来实现节点的遍历，避免了递归带来的栈溢出风险。该算法首先将根节点压入栈中，然后在遍历过程中根据当前节点和前一节点的关系决定下一步操作，最终返回节点值的后序遍历结果。

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Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ArrayList<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        if(root == null) return result ;
        stack.push(root);
        TreeNode prev = null;
        while(!stack.isEmpty()){
           TreeNode curr = stack.peek();
           // traverse down
           if(prev == null || curr == prev.left || curr == prev.right){
               if(curr.left != null) stack.push(curr.left);
               else if(curr.right != null) stack.push(curr.right);
           }// traverse up
           else if(curr.left == prev){
               if(curr.right != null) stack.push(curr.right);
           }else{
               result.add(curr.val);
               stack.pop();
           }
           prev = curr;
        }
        return result;
    }
}