Binary Tree Inorder Traversal

ID: 94; easy

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LintCode 炼码

Solution 1 (Go)

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderTraversal(root *TreeNode) []int {
    result := make([]int, 0)
    inorderHelper(root, &result)
    return result
}

func inorderHelper(root *TreeNode, res *[]int) {
    if root != nil {
        inorderHelper(root.Left, res)
        *res = append(*res, root.Val)
        inorderHelper(root.Right, res)
    }
}

Solution 2 (Java)

/**
 * 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: A Tree
     * @return: Inorder in ArrayList which contains node values.
     */
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        inorderTraversalHelper(root, result);
        return result;
    }

    private void inorderTraversalHelper(TreeNode root, List<Integer> result) {
        if (root == null) return;
        inorderTraversalHelper(root.left, result);
        result.add(root.val);
        inorderTraversalHelper(root.right, result);
    }
}

Notes

• Recursion / Divide and conquer

Solution 3 (Java)

/**
 * 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: A Tree
     * @return: Inorder in ArrayList which contains node values.
     */
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode curr = root;
        
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            result.add(curr.val);
            curr = curr.right;
        }

        return result;
    }
}

Notes

• Traversal