Leetcode: Binary Tree Postorder Traversal

最新推荐文章于 2022-07-02 16:50:24 发布

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本文介绍了一种不使用递归的方法来完成二叉树节点后序遍历的算法实现。通过栈结构辅助，该算法可以有效地追踪节点访问顺序，并避免了递归可能导致的栈溢出等问题。

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> res = new ArrayList<Integer>();
        if(root == null)
            return res;
        TreeNode pre = null;
        TreeNode cur = null;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.isEmpty()){
            cur = stack.peek();
            if(pre == null || (pre != null && (pre.left == cur || pre.right == cur))){
                if(cur.left != null)
                    stack.push(cur.left);
                else if(cur.right != null)
                    stack.push(cur.right);
            } else if (cur.left == pre){
                if(cur.right != null)
                    stack.push(cur.right);
            } else {
                res.add(stack.pop().val);
            }
            pre = cur;
        }
        return res;
    }
}