112. Path Sum

题目：

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.

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 Tree Depth-first Search 

链接：http://leetcode.com/problems/path-sum/ 

一刷，DFS

class Solution(object):
    def hasPathSum(self, root, sum):
        if not root:
            return False
        if not root.left and not root.right and sum == root.val:
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)

2/17/2017, Java, performance only bit 2%...

用类似前序遍历的方法。一定要注意什么时候需要pop，每次入栈的都是什么值。

注意叶节点是在stack的peek上，空节点不在。

recursive方法远好于iterative算法

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        TreeNode node = root;
        TreeNode previous = null;
        Stack<TreeNode> stack = new Stack<>();
        int value = sum;
        stack.push(root);
        value -= root.val;

        while (!stack.isEmpty()) {
            node = stack.peek();
            if (previous == null || previous.left == node || previous.right == node) {
                if (node.left == null && node.right == null) {
                     if (value == 0) return true;
                     else {
                         previous = stack.pop();
                         value += previous.val;
                     }
                } else if (node.left != null) {
                    stack.push(node.left);
                    value -= node.left.val;
                    previous = node;
                } else if (node.right != null) {
                    stack.push(node.right);
                    value -= node.right.val;
                    previous = node;
                }
            } else if (previous == node.left) {
                if (node.right != null) {
                    stack.push(node.right);
                    value -= node.right.val;
                    previous = node;
                } else {
                    previous = stack.pop();
                    value += previous.val;
                }

            } else if (previous == node.right) {
                previous = stack.pop();
                value += previous.val;
            }
        }
        return false;
    }
}

5/8/2017

算法班

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        if (root.left == null && root.right == null && root.val == sum) return true;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}