print out the path from one node to another in binary tree

最新推荐文章于 2023-10-09 02:15:42 发布

原创最新推荐文章于 2023-10-09 02:15:42 发布 · 784 阅读

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二叉树 (binary tree) 专栏收录该内容

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本文介绍了一种在二叉树中寻找从根节点到指定节点路径的方法，并提供了一个具体的实现示例。通过递归遍历的方式，可以找到任意两个节点的路径。

Question:

You are given one binary tree, find the path from one node to another.

public static LinkedList<Integer> printPath(Node root, Node n1, Node n2) {
		Node temp = root;
		LinkedList<Integer> rootToN1 = new LinkedList<Integer>();
		LinkedList<Integer> rootToN2 = new LinkedList<Integer>();

		pathFromRootToNode(temp, n1, new ArrayList<Integer>(), rootToN1);
		pathFromRootToNode(root, n2, new ArrayList<Integer>(), rootToN2);
		int prev = 0;;
		while (true) {
			if (rootToN1.size() > 0 && rootToN2.size() > 0 && rootToN1.get(0) == rootToN2.get(0)) {
				prev = rootToN1.get(0);
				rootToN1.remove();
				rootToN2.remove();
			} else {
				rootToN2.add(0, prev);
				for (Integer i : rootToN1) {
					rootToN2.add(0, i);
				}
				break;
			}
		}
		return rootToN2;		
	}
	// the path from root the a specific node
	public static void pathFromRootToNode(Node root, Node n, ArrayList<Integer> stack, LinkedList<Integer> list) {
		if (root == null) return;
		stack.add(root.value);
		if (root == n) {
			list.addAll(stack);
		}
		
		if (root.leftChild != null) {
			pathFromRootToNode(root.leftChild, n, stack, list);
		}
		if (root.rightChild != null) {
			pathFromRootToNode(root.rightChild, n, stack, list);
		}
		// after searching the left part, we need to remove the added value
		stack.remove(stack.size() - 1);
	}

class Node {
    Node leftChild = null;
    Node rightChild = null;
    int value;

    Node(int value) {
        this.value = value;
    }
}