Binary Tree Inorder Traversal

本文介绍两种实现二叉树中序遍历的方法:递归法和迭代法。递归法通过先遍历左子树再访问根节点最后遍历右子树的方式完成;迭代法则利用栈来保存节点,不断将当前节点的左子节点压入栈中,并将最深的左子节点弹出,访问其值,然后转向右子节点。

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1. Recursive

Straight Forward, Add left -> root -> right

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> rst = new ArrayList<Integer>();
        if(root == null) {
            return rst;
        }
        traversal(rst, root);
        return rst;
    }
    
    private void traversal(ArrayList<Integer> rst, TreeNode root) {
        if (root == null) {
            return;
        }
        traversal(rst, root.left);
        rst.add(root.val);
        traversal(rst, root.right);
    }
}

 

2. Iterator

1. Create a node to present current node.

2. If current node is not null and stack is not empty

   1) keep adding all the left child to the stack.

   2) move the current node to the top of the stack

   3) pop the top node and add it to the result list

   4) let the current node to be the right child.

 

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> rst = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode curNode = root;
        while (curNode != null || !stack.empty()) {
            //reach deepest left child
            while (curNode != null) {
                stack.push(curNode);
                curNode = curNode.left;
            }
            curNode = stack.peek();
            stack.pop();
            rst.add(curNode.val);
            curNode = curNode.right;
        }
        return rst;
    }
}

 

转载于:https://www.cnblogs.com/codingEskimo/p/6536866.html

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