【LeetCode】Path Sum ---------LeetCode java 小结
Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22
,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
思路:深度遍历(后续遍历)。每当访问到叶子节点时,计算栈内的元素,也就是从根节点到叶子节点的路径。
public boolean hasPathSum(TreeNode root, int sum) {
Stack<TreeNode> stack = new Stack<>();
int total = 0;
if (root == null)
return false;
TreeNode qNode=root;
while (root != null) {
while (root.left != null) {
stack.push(root);
total += root.val;
root = root.left;
}
while (root != null && (root.right == null || root.right == qNode)) {
if(root.right==null&&root.left==null){//访问到叶子节点
if(total+root.val==sum){
return true;
}
}
qNode = root;// 记录上一个已输出节点
if (stack.empty())
return false;
root = stack.pop();
total-=root.val;
}
stack.push(root);
total+=root.val;
root = root.right;
}
return false;
}
这个题目中,叶子结点不是非常清楚,考虑这种情况
1
/
2
结果为1,应该是正确的,存在path。
在此回顾一下二叉树的遍历(参考http://maizi2011.iteye.com/blog/938749)
/** 递归实现前序遍历 */
protected static void preorder(BTNode p) {
if (p != null) {
visit(p);
preorder(p.getLeft());
preorder(p.getRight());
}
}
/** 递归实现中序遍历 */
protected static void inorder(BTNode p) {
if (p != null) {
inorder(p.getLeft());
visit(p);
inorder(p.getRight());
}
}
/** 递归实现后序遍历 */
protected static void postorder(BTNode p) {
if (p != null) {
postorder(p.getLeft());
postorder(p.getRight());
visit(p);
}
}
/** 非递归实现前序遍历 */
protected static void iterativePreorder(BTNode p) {
Stack<BTNode> stack = new Stack<BTNode>();
if (p != null) {
stack.push(p);
while (!stack.empty()) {
p = stack.pop();
visit(p);
if (p.getRight() != null)
stack.push(p.getRight());
if (p.getLeft() != null)
stack.push(p.getLeft());
}
}
}
/** 非递归实现后序遍历 */
protected static void iterativePostorder(BTNode p) {
BTNode q = p;
Stack<BTNode> stack = new Stack<BTNode>();
while (p != null) {
// 左子树入栈
for (; p.getLeft() != null; p = p.getLeft())
stack.push(p);
// 当前节点无右子或右子已经输出
while (p != null && (p.getRight() == null || p.getRight() == q)) {
visit(p);
q = p;// 记录上一个已输出节点
if (stack.empty())
return;
p = stack.pop();
}
// 处理右子
stack.push(p);
p = p.getRight();
}
}
/** 非递归实现中序遍历 */
protected static void iterativeInorder(BTNode p) {
Stack<BTNode> stack = new Stack<BTNode>();
while (p != null) {
while (p != null) {
if (p.getRight() != null)
stack.push(p.getRight());// 当前节点右子入栈
stack.push(p);// 当前节点入栈
p = p.getLeft();
}
p = stack.pop();
while (!stack.empty() && p.getRight() == null) {
visit(p);
p = stack.pop();
}
visit(p);
if (!stack.empty())
p = stack.pop();
else
p = null;
}
}
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